Macronutrients is the most readily available for energy production

Answers

Answer 1

Answer:

The correct answer is - true.

Explanation:

All the macronutrients carbohydrates, fats, and protein all three produce energy. Carbohydrates are the most preferred source of energy for the human body as it is the macronutrient that your system most requires.

The human body easily breaks down most carbohydrates and provides a significant amount of energy. The energy-providing process is called cellular respiration starts with glucose as a substrate. Glucose is the simplest carbohydrate.


Related Questions

Suggest how whooping cough spreads from person to person​

Answers

Answer:

People with pertussis usually spread the disease to another person by coughing or sneezing or when spending a lot of time near one another where you share breathing space. Many babies who get pertussis are infected by older siblings, parents, or caregivers who might not even know they have the disease.

Explanation:

Answer:

Explanation:People with pertussis usually spread the disease to another person by coughing or sneezing or when spending a lot of time near one another where you share breathing space. Many babies who get pertussis are infected by older siblings, parents, or caregivers who might not even know they have the disease

The difference between active transport and passive transport is that a. concentration gradients are involved in one and not in the other. b. glycolipids play a role in one and not in the other. c. one requires expenditure of energy by the cell and the other does not. d. ions are transported into and out of the cell by one process and not by the other.

Answers

Answer:

D) ions are transported into and out of the cell by one process and not by the other

Glycolysis occurs in
1) mitochondria
2) cytoplasm
3) ER
5) Plastids​

Answers

Cytoplasm

Cytoplasm is a jelly-like substance between the nucleus and the cell membrane.

Various cell organelles like ribosomes, mitochondria, endoplasmic reticulum, etc. are suspended in the cytoplasm.

It helps in exchange and storage of substances among cell organelles.

Most of the metabolic reactions also occur inside cytoplasm.

What is the biggest part of the brain?

Answers

Answer:

Cerebrum

Explanation:

The cerebrum is divided into two hemispheres or halves and is the biggest portion of the brain. The cerebrum is in charge of voluntary movement, speech, intellect, memory, emotion, and sensory processing, among other things.

OAmalOHopeO

For each of the following structures, first indicate its function in the fetus; then, note its fate (what happens to it or what it is converted to after birth).

a. Umbilical artery
b. Umbilical vein
c. Ductus venosus
d. Ductus arterious
e. Foramen ovale

Answers

Answer:

1. Functions:

a. Umbilical artery >> carries deoxygenated blood from the fetus to the placenta

b. Umbilical vein >> transports oxygenated blood from the placenta to the fetus

c. Ductus venosus >> allows oxygenated blood from the placenta to bypass the liver

d. Ductus arterious >> allows most of the blood from the right ventricle to bypass the fetus's non-functioning lungs

e. Foramen ovale >> oxygenated blood from the umbilical vein to bypass the pulmonary circulation

2. After the bird:

1. Umbilical artery >> medial umbilical ligament

2. Umbilical vein >> round ligament of the liver

3. Ductus venosus  >> ligamentum venosum

4. Ductus arteriosus >> ligamentum arteriosum

5. Foramen ovale >> fossa ovalis

Explanation:

The umbilical artery is a paired artery localized in the abdominal and pelvic regions, which carries deoxygenated blood from the fetus to the placenta through the umbilical cord. The medial umbilical ligament is the obliterated part of the umbilical artery that arises from the internal iliac arteries. In utero, the umbilical arteries carry waste products back to the placenta, whereas the umbilical vein carries oxygenated blood from the placenta to the fetus. The round ligament of the liver (also known as ligamentum teres hepatis) is a remnant of the umbilical vein that  exists in the embryonic stage, it connects the left lobe of the liver to the umbilicus. The ductus venosus is a slender shunt that allows oxygenated blood from the placenta to bypass the liver, it connects the intra-hepatic portion of the umbilical vein to the inferior vena cava. The ligamentum venosum is an extrahepatic, slender, and fibrous remnant of the fetal ductus venosus that travels between the left portal vein and the inferior vena cava. The ductus arteriosus is a fetal artery that connects the aorta to the pulmonary artery. The ligamentum arteriosum is a nonfunctional vestige of the ductus arteriosus, it is attached to the superior surface of the pulmonary trunk. The foramen ovale is an oval-shaped, small, opening in the wall (septum) between the two upper chambers of the heart. The fossa ovalis is a vestige stricture of the foramen ovale of the embryonic heart, which forms a depression in the right atrium of the heart.

Part 1 of 1 -
Question 9 of 10
10 Points
When DNA is copied, sometimes there are mistakes. Approximately how often does this
happen?
O A. There aren't any mistakes.
OB. 1 in a billion bases.
OC. 1 in a million bases.
OD. 1 in a trillion bases.
Reset Selection

Answers

Answer:

D. 1 in a trillion bases

Explanation:

A mutagen agent can change the genetic information of organisms increasing mutations over the natural level. Mutagens cause changes in the bases, and pairing bases, that compose DNI strands.

A mistake in the process of DNI copy during cell division might cause genetic changes in daughter cells. Defects DNI replication might be inherited if it occurs in germinal cells. But it can also cause many significant epigenetic changes.

Many of these changes can be detected on time by enzymes such as DNI polymerase. This enzyme can correct these mistakes or at least some of them, moving from 3´to 5´direction, and eliminating the mistakes.

The highly effective replication system, together with the action of enzymes, makes it rare to occur a mistake in DNI replication. Generally speaking, the mistaken rates in DNI replication are very low, meaning that only one in a trillion times occurs a mistaken DNI copy.

Which structure transports urine to the bladder by peristaltic action?

Answers

Answer:

The muscular layer of the ureter consists of longitudinal and circular smooth muscles that create the peristaltic contractions to move the urine into the bladder without the aid of gravity.

Answer:

The ureter

Explanation:

The ureter is a long thin tubular structure 10-12 inches long which carries urine produced in the kidney to the bladder. The urine is transported by a process called peristalsis. The ureter actively propels urine from the kidney down into the bladder.

2. At what temperature did the prodigiosin-producing genes express in the S. marcescens culture? From the experiment you conducted in this lab, what evidence can you provide to support your claim?

Answers

Answer:

The temperature is the key factor for prodigiosin production. It has been shown that , S. marcescens can produce this kind of pigment at about 25 °C, which however could not produce the pigments at elevated temperatures, especially till 37 °C

Explanation:

Histones are essentially identical in sequence/structure in all eukaryotic organisms from yeast to plants to animals. What does this say about the biophysical properties of DNA-packaging and the evolution of eukaryotic organisms

Answers

Answer:

It indicates that core histone genes were present in the last common ancestor of yeasts, plants, and animals

Explanation:

Histones are highly basic proteins that can strongly interact with DNA, which is packaged into nucleosomes, the basic structural and functional unit of chromatin. Each nucleosome is composed of approximately 147 base pairs of DNA wrapped around a core of eight histone proteins (two copies of four types of histones H3, H4, H2A, H2B). These core histones are evolutionarily conserved across eukaryotic kingdoms in terms of sequence and structure. Therefore, DNA-packaging into nucleosomes is considered a constraint for the evolution of core histones. Moreover, the presence of conserved core histones in eukaryotic kingdoms (e.g., yeast, plant, and animal kingdoms) is strong evidence that histone-mediated DNA packaging was presumably present in the last common ancestor of eukaryotic genomes.

A sample from a stock of a bacterial colony in liquid media was diluted by a factor of 106, and 2 ml of this dilution was spread on a Petri dish of solidified media. 56 colonies were observed. What was the concentration of bacteria of the initial stock?

Answers

Answer:

28 × 10⁶ colonies/ml

Explanation:

Let C be the concentration of bacterial in the initial stock. When it is diluted by a factor of 10⁶, the new concentration is C' = C/10⁶.

When 2 ml of this concentration is spread on a Petri dish of solidified media, 56 colonies were produced. The number of colonies, n after spreading the 2 ml of C' is C' × 2 ml = 2C' = 2C/10⁶.

So, n = 2C/10⁶.

Since the number of colonies after spreading on a Petri dish of solidified media is 56, n = 56 colonies.

So, 2C/10⁶ = 56

Making C subject of the formula, we have

C = 56 × 10⁶/2

C = 28 × 10⁶ colonies/ml

So, the initial concentration of bacteria is 28 × 10⁶ colonies/ml

Select the logical fallacy used in this statement: If we ban semiautomatic weapons, it won't stop until handguns and rifles are banned as well, and then the criminals will have all the guns.

Select one:
a.
Slippery Slope
b.
Ad Populum
c.
Ad Hominem
d.
There is no fallacy in this statement

Answers

Answer:

A- Slippery Slope

Explanation:

A slippery slope fallacy is when someone claims, without proper evidence, that an action will lead to an often catastrophic consequence via a series of events.

Here, the statement claims that if semi-automatics are banned, so will handguns and rifles, however, doesn't provide any evidence that this is the case.

Which type of seedless plant has a complex leaf arrangement off a vein?
a. java moss
b. club moss
c. ferns
d. horstails

Answers

B. Club moss

Explanation:

This is because club moss is an seedless evergreen plants that have scale-like leaves.

Can you plz mark me as brainliest!!!

answer : club moss

explanation: Because they have vascular tissue, seedless vascular plants

are often larger than nonvascular plants. Vascular tissue is spe-

cialized to transport water to all of the cells in a plant.

Which of the following would provide the best evidence that species A and species B have a common ancestor?
O A. The limbs of species A perform a function that is similar to the limbs of species B, but they have a different structure.
OB. Species A has limbs, and species B does not have limbs.
O c.
The limbs of species A and species B have a different structure and function.
OD
The limbs of species A have a similar structure to the limbs of species B, but they perform a different function.

Answers

Answer:

I believe the answer is D

Explanation:

Answer:

B

Explanation:

The answer is B on Plato. Just took the test and got it right.

Traditionally, the classification of fungi has been based on the nature of sexual stages of the life cycle. For Penicillium, however, no sexual stages of the life cycle have been observed. Without evidence from sexual stages, speculate about other possible sources of evidence that scientists may use in classification.

Answers

Answer:

Without the evaluation of sexual stages, fungi can be classified according to the type of colony they present, the speed of growth, the formation of pigments and the type of coloration.

Explanation:

Although the observation of sexual stages is extremely efficient for the classification of fungi in the laboratory. This type of analysis is not always possible to be carried out. In that case, scientists need to find other methods that allow for the classification of fungi. These methods are carried out with the help of a microscope, where scientists observe the morphology of the fungi and are able to classify them according to the type of colony they present, the speed of growth, the formation of pigments and the type of coloration.

Considering your knowledge of carbohydrates, evaluate the use of chitin as a component of health foods.

Answers

Answer:

Carbohydrates may be defined as energy-rich foods such as sugars and starches. if consumed in excess or not properly metabolized in the body, carbohydrates may lead to obesity and which may also lead a person to severe number of diseases.

Chitins are components of the exoskeletons of foods such as shrimps, crabs and snails, and insects.

Chitin's use as a component in healthy foods is based on its health benefits.

For example it promotes weight loss , prevents obesity, relieving constipation and preventing inflammation associated with refined carbohydrates, cookies and candies

Match the following description with the appropriate type of respiration:

a. occurs in the mitochondria
b. resting muscles depend on this type of respiration
c. lactic acid builds up in muscle fibers
d. rapidly produces ATP for short time periods
e. produces large quantities of ATP but takes longer to synthesize

1. anaerobic respiration
2. aerobic respiration

Answers

Answer:

The correct answer is -

1. c, and d.

2. a, b, and e.

Explanation:

Anaerobic respiration is the short and quick way of producing energy, however, it produces less amount of energy than aerobic respiration and produces lactic acid as a byproduct. It takes place in the cytoplasm of the cell and working or acting muscles depend on such respiration.

Aerobic respiration is the main and important respiration process that takes place in mitochondria and it takes time to produce ATPs that are much more than anaerobic respiration. Resting muscles get energy by this type of respiration.

what are the effects of under secretion of steroid hormones​

Answers

Answer:

Anabolic steroids are related to testosterone, the major male hormone. Abusing the hormone can lead to physical and psychological side effects. Problems include breast development and hair loss among men, and facial hair growth, menstrual problems, and a deepened voice in women.

Issued in 1974, 45 CFR 46 raised to regulatory status:
A) The 1974 National Research Act
B) The Nuremberg Code
C) Kefauver-Harris Drug Amendments to the Federal Food, Drug & Cosmetics Act
D) US Public Health Service Policy

Answers

Hello!

The answer is D, The U.S Public Health Service Policy. Its main purpose was providing protection for human subjects for research work which was conducted by federal agencies. You can read more about it on HHS.gov, as it is a very interesting regulation.

I hope this helps! :)

Issued in 1974, 45 CFR 46 raised to regulatory status US Public Health Service Policy. Option D

What is the regulatory status?

The restrictions known as the Common Rule for the protection of human subjects in research done by or supported by federal agencies are outlined in Title 45 of the Code of Federal restrictions, Part 46. These laws define moral principles and requirements for the protection of research subjects who are being used as human subjects.

The US Public Health Service (PHS) published the policy that would eventually become 45 CFR 46 in 1974. This policy established standards for the examination and approval of research involving human subjects as well as the legal foundation for the protection of those individuals. Informed consent, weighing risks and benefits, and the creation of Institutional Review Boards (IRBs) to regulate research techniques were all adopted.

Learn more about regulatory status:https://brainly.com/question/30479939

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Give reason. Mosquitoes and housefly are placed in the phylum arthropoda​

Answers

Mosquitoes and Housefly are grouped under the phylum Arthropoda because they both posses jointed legs (or appendages).

Animals that have jointed appendages are called Arthropods (i.e animals with jointed legs). They are triploblastic (i.e they have 3 germ layers) and exhibit bilateral symmetry. They are both six-legged arthropods sub-grouped under the class Hexapoda (coined from the word "hexa" meaning six and "Poda" meaning leg). Their body are divided into three parts; head, thorax and abdomen.

Learn more:

https://brainly.com/question/2244172?referrer=searchResults

what is The Catalys?​

Answers

Answer:

A catalyst is a chemical substance that speeds up the rate of a chemical reaction at any given conditions.

56:25
If blood is in short supply, which blood type would be the most beneficial to have on hand if someone needed a blood transfusion?
O+
O–
AB+
AB–

Answers

I think it’s o+ if not try o- sorry if this was no help

A genetically heterogeneous population of rice has a mean in the number of days to maturation of 30. Selection for decreased period of maturation is carried out for one generation. The average period to maturation among the plants selected as parents for the next generation is 25 days. F1 plants mature on average in 27 days. Estimate the narrow sense heritability.

Answers

Answer:

h² = 0.6

Explanation:

Before answering the question, we need to know a few concepts.

Artificial selection is the selecting practice of a specific group of organisms in a population -that carry the traits of interest- to be the parents of the following generations.

Parental individuals carrying phenotypic values of interest are selected from the whole population. These parents interbreed, and a new generation is produced.

The selection differential, SD, is the difference between the mean value of the trait in the population (X₀) and the mean value of the parents, (Xs). So,  

SD = X₀ - Xs

Heritability in the strict sense, h², is the genetic component measure to which additive genetic variance contributes. The heritability might be used to determine how the population will respond to the selection done, R.  

h² = R/SD

The response to selection (R) refers to the metric value gained from the cross between the selected parents. R can be calculated by multiplying the heritability h², with the selection differential, SD.  

R = h²SD  

R also equals the difference between the new generation phenotypic value (X₁) and the original population phenotypic value (X₀),  

R = X₀ - X₁

-------------------------------------------------------------------------------------------------------------

Now that we know these concepts and how to calculate them, we can solve the proposed problem.

Available data:  

You are selecting rice´s decreased period of maturation. The population of rice has a mean maturation time of 30 days → X₀ Parental selected average period to maturation is 25 days → Xs F1 plants mature on average in 27 days → X₁ N arrow sense Heritability → h²

According to what we sow previously, we need to find out the value of h².

We know that h² = R/SD, so we need to get R and SD first.  

R = X₁ - X₀

R = 27 - 30

R = -3

SD = Xs -  X₀  

SD = 25 - 30

SD = -5

Knowing this, we can calculate h²

h² = R/SD

h² = -3/-5  

h² = 0.6

 

water can act as either a(n)__or a​

Answers

Explanation:

Water can act as an acid and a base. As an acid, water donates H+, the hydrogen ion. As a base, water donates OH-, the hydroxide ion

hope this helps you

have a nice day:)

Given that the intracellular concentration of potassium is 150 mEq/L, how would the potassium equilibrium potential be affected if the extracellular concentration of potassium is changed from 5.0 to 3.5 mEq/L

Answers

Answer:

The potassium equilibrium potential would increase, meaning that more K+ would be leaving the cell.

Explanation:

Let us assume that the only ion transported through the cell membrane is K+. We need to use the Nernst equation to know the destiny of the ion.

Nernst equation:

E = 58 millivolts/z. [Log₁₀ (C-out/C- in)

Where,    

• E = Equilibrium potential

• 58 millivolts/z = Constant

• z = Ion charge + positive or negative symbol

• C-out = Ion concentration out of the cell

• C-In = Ion concentration inside the cell

By convenience, in the Nerts equation, the bigger concentration value corresponds to the numerator and the smaller concentration value to the denominator.

Now let us see the provided values,

• z = Ion charge + positive or negative symbol ⇒ +1 ⇒ K+

• C-out = Ion concentration out of the cell ⇒ 5 mEq/L

• C-In = Ion concentration inside the cell ⇒ 150 mEq/L

E = 58 millivolts/z. [Log₁₀ (Ion in/Ion out)

E = 58 millivolts/+1. [Log₁₀ (150 mEq/L / 5 mEq/L)

E = 58 millivolts (Log₁₀ 30)

E = 58 millivolts (1.477)

E = 85.67 millivolts

85.7 mV is the absolute value of equilibrium potential.

E = 58 millivolts/z. [Log₁₀ (Ion in/Ion out)

E = 58 millivolts/+1. [Log₁₀ (150 mEq/L / 3.5 mEq/L)

E = 58 millivolts (Log₁₀ 42.85)

E = 58 millivolts (1.63)

E = 94.65 millivolts

94.7 mV is the absolute value of equilibrium potential.

If the extracellular concentration of potassium is changed from 5.0 to 3.5 mEq/L, there will be an increase in the membrane potential from 85.7 to 94.7 mV. The increase in the equilibrium potential will result in more potassium diffusing out of the cell, turning the cell interior less positive than before.  

The potassium equilibrium ability might increase, which means that greater K+ might be leaving the cell.

Let us expect that the handiest ion transported via the cell membrane is K+. We want to apply the Nernst equation to recognize the future of the ion.

Nernst equation:

[tex]E = 58 millivolts/z. [Log₁0 (C-out/ -in)[/tex]

Where,

E = Equilibrium ability.58 millivolts/z =Constant.z=lon charge + advantageous or terrible symbol. C - out = Ion awareness out of the cell. C-ln= ion awareness in the cell.

For convenience, withinside the Nerts equation, the larger awareness fee corresponds to the numerator and the smaller awareness fee to the denominator. Now allow us to see the supplied values,

[tex]z=lon charge + effective or terrible ⇒+1 ⇒ K+\\C - out = lon awareness out of the cell ⇒5 mEq/L\\C-ln= lon awareness withinside the cell ⇒150 mEq/LE = 58 millivolts/+ 1.[Log 10 ( 50mEq / L / 5mEq/L)\\E = 58millivolts (Log30)\\E = 58 millivolts (1.477)[/tex][tex]E = 85.67 millivolts\\85.7 mV is absolutely the fee of equilibrium capability.\\E = 58 millivolts/z. [Log10 (lon in/lon out)\\E =58 millivolts/+1. [Log 10 (a 100and 50 mEq/L / 3.five mEq/L)\\E =58 millivolts (Log10 42.85)\\E = 58 millivolts (1.63)\\E = 94.65 millivolts94.7 mV is absolutely the fee of equilibrium capability.[/tex]

What happens if the extracellular attention of potassium is modified from 5.0 to 3.5?

If the extracellular attention of potassium is modified from 5.0 to 3.5mEq/L, there can be a growth withinside the membrane capability from 85.7 to 94.7 mV. The growth withinside the equilibrium capability will bring about extra potassium diffusing out of the cell, turning the cell indoors much less high quality than before.

Thus it is clear from this that the potassium equilibrium potential is affected.

To know more about the potassium equilibrium refer to the link :

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Which type of energy refers to the sum of potential and kinetic energies in the particles of a substance?

Answers

Answer:

The answer is Internal energy

KE + PE = IE

Explanation:

The sum of potential and kinetic energy is refers to mechanical energy which is expressed by motion  

Answer:C for edge (internal energy)

Explanation:

Which is a compound that allows plants to get nitrogen from the nitrogen cycle?

Answers

Answer:

Plants can use ammonia as a nitrogen source. After ammonium fixation, the ammonia and ammonium that is formed will be transferred further, during the nitrification process. Aerobic bacteria use oxygen to convert these compounds.

Lainey is looking for a new apartment and her realtor keeps calling her with new listings. The calls only take a few minutes, but a few minutes here and there are really starting to add up. She's having trouble concentrating on her work. What should Lainey do? a) Tell her realtor she can only receive text messages O b) Limit the time spent on each call O c) Turn off her phone until she is on a break O di Call her realtor back when customers won't see her on the phone

Answers

Answer:

c) Turn off her phone until she is on a break

Explanation:

If she does option "a" then her phone will still keep ringing with text message alerts. Option " b" will still be consuming her work time and option "d" can't be because it'll be too late by then so only option c makes sense.

explain how the tissue of the esophagus and tissue of the trachea can be differentiate

Answers

Answer:

Trachea: It is the wind pipe — making it a part of the respiratory system

Esophagus: It is the food pipe — making it a part of the digestive system

Trachea: It is shorter, 10–11 centimeters. It connects upper airway to the lungs

Esophagus: It is longer, 25 centimeters. It connects mouth to the stomach

Trachea: It is cartilaginous, made of C-shaped semicircular cartilages. They give it structural stability and prevents it from collapsing

Esophagus: It is muscular. It contracts in a wave-like motion through it’s length to propel food from mouth to stomach a.k.a swallowing

Trachea: It’s opening is protected by Epiglottis, a flap like structure, to prevent food from accidental entering the air passage. It prevents choking

Esophagus: It’s opening is protected by two sphincters. They are muscular rings that constrict to close the esophagus off when food is not being swallowed.

Trachea: It has 2 portions — cervical and thoracic i.e. neck and chest portions.

Esophagus: It has 3 portions — cervical, thoracic and abdominal i.e. neck, chest and stomach portions.

Kevin's supervisor, Jill, has asked for an update on today's sales, Jill is pretty busy moving back and forth between different store locations. How can Kevin most effectively deliver an update to her ? a) Call with a quick update Ob ) Send a detailed text message c ) Book a one-hour meeting for tomorrow morning d) Send a detailed email

Answers

Answer:

d) Send a detailed email

Explanation:

Send a detailed email is the best way to deliver an update to her because in the email he can send all the information in detail form which can satisfy his owner. He can't call or message because it takes too much time to provide information so email is the best way to provide information. Booking a one-hour meeting is not worth it and the reason for this is that there is no big presentation which takes one hour of description one email is enough for it.

100 POINTS!!!!!!
The theories surrounding the formation of our solar system are based on many

biased opinions
incorrect facts
non-testable data
scientific investigations

Answers

scientific investigations is the correct thing

Answer:

"scientific investigations."

Other Questions
Read the hypothetical situation below and answer the question that follows.When Michaels employer finds out that he voted in an election, he is fired from his job immediately.Which of the following voting obstacles has Michael faced?A.poll taxingB.literacy testingC.voting fraudD.intimidation and violence You want to retire exactly 30 years from today with $1,950,000 in your retirement account. If you think you can earn an interest rate on 10.07 percent compounded monthly, how much must you deposit each month to fund your retirement an umbrella is made by stitching 10 triangular pieces of cloth of two different colours where each pieces measuring 48cm,48cm,and 30cm. find cost of required to make umbrella at rate of 5paisa per square cm WM "Please write the molecular formula for the following by criss cross method. 1. Calcium nitrate 2. Sodium bicarbonate 3. Magnesium nitrite 4. Sodium chloride 5. Hydrogen sulphate 6. Hydrogen chloride T-Aluminium bisulphate 8 Aluminium chloride 9. Aluminium sulphate 10. Hydrogen Nitrate Mr. and Mrs. Nunez attended one of your sales presentations. Theyve asked you to come to their home to clear up a few questions. During the presentation, Mrs. Nunez feels tired and tells you that her husband can finish things up. She goes to bed. At the end of your discussion, Mr. Nunez says that he wants to enroll both himself and his wife. What should you do? find m Gi S l tp cc gi tr nguyn ca tham s m trong khong (20;20) bt phngtrnh +2{m+2n -m+36]x+m+10-mnghim ng vi mi xe s phn t ca S. TnhA25.B20.C19.=U calculate the length of wire. Which of the following best describes the journal entry to record the withdrawal of raw materials from the storeroom for use as direct and indirect materials in production?a. Debit Work in Process, debit Manufacturing Overhead, and credit Raw Materials. b. Debit Work in Process and credit Raw Materials. c. Debit Manufacturing Overhead and credit Raw Materials. d. Debit Work in Process, debit Manufacturing Overhead, and credit Direct Materials. write any five advantage of good company for disadvantage of bad company ? What are the solutions of x2 - 2x +5 = 0 Following are selected accounts for a manufacturing company. For each account, indicate whether it will appear on a budgeted income statement (BIS) or a budgeted balance sheet (BBS). If an item will not appear on either budgeted financial statement, write it NA.a. Sales b. Administrative salaries paidc. Accumulated depreciationd. Depreciation expensee. Interest paid on bank loanf. Cash dividends paidg. Bank loan owedh. Cost of goods sold what are the effects of under secretion of steroid hormones Science, Geology, Class 81. choose the correct answer ( CTCA )a. what makes the soil fertile?i. rock particles ii. humusiii. sand particlesiv. silt particles Natalie is 4 years old. When she buttons her shirt, she talks to herself and describes the steps. This helps her in self-regulating and guiding her behavior. In the context of cognitive and physical development in early childhood, this scenario illustrates the concept of Nyasia had 16 M&Ms on her desk. She had 2 groups of M&Mswith x amount in each, and another group with 4 M&Ms. Howmany M&Ms were in the identical groups? Jagadison Co. leases computer equipment to customers under sales-type leases. The equipment has no residual value at the end of the lease and the leases do not contain purchase options. Jagadison desires a return of 11% interest on a four-year lease of equipment with a fair value of $795,564. The present value of an annuity due of $1 at 11% for four years is 3.444. What is the total amount of interest revenue that Jagadison will earn over the life of the lease?a. 128436b. 198891c. 231000d. 350048 Landon Stevens is evaluating the expected performance of two common stocks, Furhman Labs, Inc., and Garten Testing, Inc. The risk-free rate is 4.4 percent, the expected return on the market is 10.6 percent, and the betas of the two stocks are 1.4 and 0.7, respectively. Stevenss own forecasts of the returns on the two stocks are 10.60 percent for Furhman Labs and 10.50 percent for Garten.Required:a. Calculate the required return for each stock.b. Is each stock undervalued, fairly valued, or overvalued? 5 (6 - x) - 3 = 5 + 4 (x + 3) When increasing production from 12,000 computers to 15,000 computers, the company's average cost of production will A. increase from $10.10 to $10.40 due to the learning-curve effect. B. increase from $16.80 to $18.90 due to the learning-curve effect. C. decrease from $10.40 to $10.10 due to diseconomies of scale. D. decrease from $10.40 to $10.10 due to the learning-curve effect. E. increase from $16.80 to $18.90 due to economies of scale.