Graph the linear equation by find
2 = 4x + y

Answer:

Pretty sure its y=4x-5 hope it helped

Step-by-step explanation:

C-x=-5

Step-by-step explanation:

-3x=-15

or,-3x x =-15

or, x=-15÷3

therefore, x=-5

Answer:

C.

Step-by-step explanation:

A P E X

(256⁴-1)

= (256-1)⁴

Using identity (a-b)⁴ = a⁴−4a³b+6a²b²−4ab³+b⁴

Let a be 256 and b be 1

Then

256⁴−4(256)³(1)+6(256)²(1)²−4(256)(1)³+(1)⁴

After solving

(256²-1)²

(a-b)² = a²-2ab+b²

256²-2×256×1+1²

= (256²-1)(256²+1)

Must click thanks and mark brainliest

Answer:

Use identity:

Consider that:

Now factorize:

Answer:

26 + 3 x 42 + 7 x -2 = 138

Step-by-step explanation:

Ok bud, first step we must convert our symbols (Makes it easier to solve)

26 + 3 x 42 + 7 x -2

* subsitutes for multiplication.

I recommend using PEMDAS at times:

1 - Parentheses

2 - Exponents and Roots

3 - Multiplication

4 - Division

5 - Addition

6 - Subtraction

Yet again your numbers were spaced out could they be exponents? if so:

3x^{42}+7x+24

Our answer would round to 24 but he equation was not put in a valid or straight forward way.

Answer:

Step-by-step explanation:

a

fdsa

Answer:

Here we only need to analyze the vertical motion of the ball.

First, because the ball is in the air, the only force acting on the ball will be the gravitational force (we are ignoring air resistance), then the acceleration of the ball is equal to the gravitational acceleration, so we have:

a(t) = -32 ft/s^2

where the negative sign is because this acceleration is downwards.

To get the velocity equation, we need to integrate over the time, so we get:

v(t) = (-32 ft/s^2)*t + V0

where V0 is the initial velocity of the ball. In this case, the initial velocity of the ball will be the velocity that the ball had when it was dropped, which should be the same as the velocity of the hot air ballon, so we have:

V0 = 12 ft/s

Then the velocity equation of the ball is:

v(t) = (-32 ft/s^2)*t + 12 ft/s

To get the position equation we integrate again:

p(t) = (1/2)*(-32 ft/s^2)*t^2 + (12 ft/s)*t + H0

Where H0 is the initial height. We know that the ball was released at the height of 1120 ft, then we have:

H0 = 1120 ft.

Then the position equation is:

p(t) = (-16 ft/s^2)*t^2 + (12 ft/s)*t + 1120ft

a)  How many seconds after its release will the ball strike the ground?

The ball will strike the ground when its position equation is equal to zero, then we need to solve:

p(t) = 0 ft = (-16 ft/s^2)*t^2 + (12 ft/s)*t + 1120ft

This is just a quadratic equation, the solutions are given by Bhaskara's formula, so the solutions for t are:

[tex]t = \frac{- 12ft/s \pm \sqrt{(12 ft/s)^2 - 4*(-16ft/s^2)*(1120 ft)} }{2*(-16ft/s^2)}[/tex]

We can simplify that to get:

[tex]t = \frac{-12ft/s \pm 268ft/s}{-32ft/s^2}[/tex]

So we have two solutions:

[tex]t = \frac{-12ft/s + 268ft/s}{-32ft/s^2} = -8s[/tex]

[tex]t = \frac{-12ft/s - 268ft/s}{-32ft/s^2} = 8.75s[/tex]

The negative solution does not make sense, then the correct solution is the positive one.

We can conclude that the ball will hit the ground after 8.75 seconds.

b)  At what velocity will it hit the ground?

We already know that the ball strikes the ground 8.75 seconds after it is released.

The velocity at it hits the ground is given by the velocity equation evaluated in that time:

v(8.75 s) = (-32 ft/s^2)*8.75s + 12 ft/s = -268 ft/s

Answer:

AC, AB, BC

Step-by-step explanation:

Given

[tex]\triangle ABC[/tex]

Required

Name the sides

(a) [tex]\angle A[/tex] and [tex]\angle C[/tex]

This represents side [tex]AC[/tex]

(b) [tex]\angle A[/tex] and [tex]\angle B[/tex]

This represents side [tex]AB[/tex]

(c) [tex]\angle C[/tex] and [tex]\angle B[/tex]

This represents side [tex]CB[/tex]

i.e. we simply bring the name of both angles together

Answer:

AC, AB, BC

Step-by-step explanation:

Nice job on getting the correct answer 144 for the first box. For anyone curious, this is found by multiplying the side lengths of the square. So 12*12 = 144.

======================================

All four sides are decreased by 5 meters. So we go from 12 meter sides to 12-5 = 7 meter sides.

The area of the square is found in the same way as earlier: We multiply the side by itself

area = side*side = 7*7 = 49 square meters

So you'll type 49 into the second box

=======================================

To find the amount of decrease, you subtract the two area values:

144 - 49 = 95 goes in the third box

This indicates that the area has decreased by 95 m^2

Answer:

hello,

a) answer: 600

b) answer: 840

Step-by-step explanation:

a)

20=2²*5,25=5²,30=2*3*5,40=2³*5

l.c.m(20,25,30,40)=2³*3*5²=8*3*25=800

b)

24=2³*3, 42=2*3*7,35=5*7

l.c.m=(24,42,35)=2³*3*5*7=840

Answer:

Part A: y = 25x + 30

Part B: $180

Step-by-step explanation:

for part A, 25 is the constant so that goes with the x, and you just add 30 because it is the extra fee. slope intercept form is y = mx + b

for part B you just plug 6 into the slope intercept equation

y = 25(6) + 30

y = 150 + 30

y = 180

$180

The advertiser failed to reject the null hypotenuse so this would mean there is not sufficient evidence at the 0.02 level of significance to say that the percentage of people who own a Rolls Royce is higher than 47%

Answer:

(a) 4 sample points

(b) See attachment for tree diagram

(c) The probability that no tail is appeared is 1/4

(d) The probability that exactly 1 tail is appeared is 1/2

(e) The probability that 2 tails are appeared is 1/4

(f) The probability that at least 1 tail appeared is 3/4

Step-by-step explanation:

Given

[tex]Coins = 2[/tex]

Solving (a): Counting principle to determine the number of sample points

We have:

[tex]Coin\ 1 = \{H,T\}[/tex]

[tex]Coin\ 2 = \{H,T\}[/tex]

To determine the sample space using counting principle, we simply pick one outcome in each coin. So, the sample space (S) is:

[tex]S = \{HH,HT,TH,TT\}[/tex]

The number of sample points is:

[tex]n(S) = 4[/tex]

Solving (b): The tree diagram

See attachment for tree diagram

From the tree diagram, the sample space is:

[tex]S = \{HH,HT,TH,TT\}[/tex]

Solving (c): Probability that no tail is appeared

This implies that:

[tex]P(T = 0)[/tex]

From the sample points, we have:

[tex]n(T = 0) = 1[/tex] --- i.e. 1 occurrence where no tail is appeared

So, the probability is:

[tex]P(T = 0) = \frac{n(T = 0)}{n(S)}[/tex]

This gives:

[tex]P(T = 0) = \frac{1}{4}[/tex]

Solving (d): Probability that exactly 1 tail is appeared

This implies that:

[tex]P(T = 1)[/tex]

From the sample points, we have:

[tex]n(T = 1) = 2[/tex] --- i.e. 2 occurrences where exactly 1 tail appeared

So, the probability is:

[tex]P(T = 1) = \frac{n(T = 1)}{n(S)}[/tex]

This gives:

[tex]P(T = 1) = \frac{2}{4}[/tex]

[tex]P(T = 1) = \frac{1}{2}[/tex]

Solving (e): Probability that 2 tails appeared

This implies that:

[tex]P(T = 2)[/tex]

From the sample points, we have:

[tex]n(T = 2) = 1[/tex] --- i.e. 1 occurrences where 2 tails appeared

So, the probability is:

[tex]P(T = 2) = \frac{n(T = 2)}{n(S)}[/tex]

This gives:

[tex]P(T = 2) = \frac{1}{4}[/tex]

Solving (f): Probability that at least 1 tail appeared

This implies that:

[tex]P(T \ge 1)[/tex]

In (c), we have:

[tex]P(T = 0) = \frac{1}{4}[/tex]

Using the complement rule, we have:

[tex]P(T \ge 1) + P(T = 0) = 1[/tex]

Rewrite as:

[tex]P(T \ge 1) = 1-P(T = 0)[/tex]

Substitute known value

[tex]P(T \ge 1) = 1-\frac{1}{4}[/tex]

Take LCM

[tex]P(T \ge 1) = \frac{4-1}{4}[/tex]

[tex]P(T \ge 1) = \frac{3}{4}[/tex]

Answer:

The rectangles are not similar

Step-by-step explanation:

In order to check whether they are similar, we need to take the ratio of their similar sides and ensure they are equal to a constant k.

Hence;

AB/PL = AD/LM = k

32/26 = 18/12 = k

16/13 = 9/6 = k

Since the scale factor is not the same, hence the rectangles are not the same.

Answer:

0

Step-by-step explanation:

Arc sec(1)=0, tan(0)=0

Answer:

The mean of these numbers is 15.68816 with a repeating 6.

Step-by-step explanation:

Answer:

[tex]\displaystyle x \approx -4.28[/tex]

General Formulas and Concepts:

Pre-Algebra

Algebra II

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle 1 = ln(x + 7)[/tex]

Step 2: Solve for x

e^1 = x+7

e - 7 = x

x = -4.28

Answer:

Step-by-step explanation:

whitch answer how do you want us to answer

[tex]\\ \large\sf\longmapsto x + 15 = 2x - 40 \\ \\ \large\sf\longmapsto x - 2x = - 40 - 15 \\ \\ \large\sf\longmapsto - x = - 55\\ \large\sf\longmapsto x = 55[/tex]

The value of x is 55 degrees.

A rhombus is a diamond-shaped quadrilateral that has all four sides equal

The properties of a rhombus are:

According to the given question.

We have a rhombus.

Since, we know that the diagonals of the rhombus bisect the angles.

Therefore,

[tex](x+15)^{o} = (2x-40)^{o}[/tex]

Solve, the above expression for x.

[tex]\implies x +15 = 2x -40\\\implies 15+40 = 2x -x \\\implies 55 = x \\0r\ x = 55^{o}[/tex]

Hence, the value of x is 55 degrees.

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Answer:

Option D

Only the equation in option D matches with the table

Answered by GAUTHMATH

The expected value of the distribution of the number of selected red balls is 0.795.

The expected value of the distribution is the mean or average of the possible outcomes.

There are 12 balls in an urn, five of which are crimson. The selection of a red ball is desired and hence considered a success.

In this case, the possible outcomes are 0, 1, 2, or 3 red balls.

To calculate the expected value, we need to find the probability of each outcome and multiply it by the value of the outcome.

The probability of selecting 0 red balls is :

[tex]$\frac{7}{12} \cdot \frac{6}{11} \cdot \frac{5}{10} = \frac{105}{660}$[/tex].

The probability of selecting 1 red ball is :

[tex]$3 \cdot \frac{5}{12} \cdot \frac{7}{11} \cdot \frac{6}{10} + 3 \cdot \frac{7}{12} \cdot \frac{5}{11} \cdot \frac{6}{10} + 3 \cdot \frac{7}{12} \cdot \frac{6}{11} \cdot \frac{5}{10} = \frac{315}{660}$[/tex].

The probability of selecting 2 red balls is

[tex]:$\dfrac{5}{12} \cdot \frac{7}{11} \cdot \frac{6}{10} + \frac{7}{12} \cdot \frac{5}{11} \cdot \frac{6}{10} + \frac{7}{12} \cdot \frac{6}{11} \cdot \frac{5}{10} = \frac{105}{660}$.[/tex]

The probability of selecting 3 red balls is

[tex]$\dfrac{5}{12} \cdot \frac{4}{11} \cdot \frac{3}{10} = \frac{15}{660}$[/tex]

The expected value is then :

[tex]$0 \cdot \frac{105}{660} + 1 \cdot \frac{315}{660} + 2 \cdot \frac{105}{660} + 3 \cdot \frac{15}{660} = \frac{525}{660} = \frac{175}{220} \approx \boxed{0.795}$[/tex]

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Answer:

y = -3x/4 + 4

Step-by-step explanation:

slope m = -3/4

-2=(-3/4)×8+b

or, b = 4

y = mx + b

y = -3x/4 + 4

Answered by GAUTHMATH

Answer:

3-4/5=2.2

hope it helps

Hi there!  

»»————-　★　————-««

I believe your answer is:  

[tex]5\frac{7}{10}[/tex]

»»————-　★　————-««  

Here’s why:  

⸻⸻⸻⸻

[tex]\boxed{\text{Calculating the answer...}}\\\\6\frac{1}{2}-\frac{4}{5} \\-------------\\6\frac{1}{2} = \frac{13}{2} \\\\\frac{13}{2} - \frac{4}{5}\\\\LCM(2,5): 10\\\\\frac{13}{2} =\frac{13*5}{2*5} =\frac{65}{10}\\\\\frac{4}{5}=\frac{4*2}{5*2}=\frac{8}{10}\\\\\frac{65}{10}-\frac{8}{10}\\\\ \frac{57}{10}\\\\\frac{57}{10}\rightarrow\boxed{5\frac{7}{10}}[/tex]

»»————-　★　————-««  

Hope this helps you. I apologize if it’s incorrect.  

Answer:

other two angle will be

82

as 82+82+16 = 180'

Step-by-step explanation:

We want to find how many miles per minute Joe runs. This means, we need to simplify the equation:

[tex]\frac{12 mi}{84 min}[/tex]

so that the number on the bottom is 1. How do we do that?

Simple.

Just divide 12 / 84

Answer:

Joe runs approx. 0.143 miles per minute.

Answer:

a rational number is a number that can be written as a fraction

I hope this helps

Step-by-step explanation:

here's the answer to your question

Answer:

the answer to your question is 7.8 (E)

The  distance from point N to LM is 7.8, 8.1 and 8.4 unit.

Perpendicular lines are those that cross at a straight angle to one another. Examples include the opposite sides of a rectangle and the steps of a straight staircase. the icon used to represent two parallel lines.

Perpendicular lines are two separate lines that cross one other at a right angle, or a 90° angle.

Given:

In ΔNOM

The Perpendicular distance is 7.8 unit

and, Hypotenuse distance id 8.1 unit

Now, In ΔNOL

The Perpendicular distance is 7.8 unit

and, Hypotenuse distance id 8.4 unit

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