The number of ounces in an unknown number of pounds. Let z= the variable.

Answer:

C

Step-by-step explanation:

We are given that:

[tex]\displaystyle \sin y^\circ = \frac{s}{8}\text{ and } \tan y^\circ = \frac{s}{t}[/tex]

And we want to find the value of:

[tex]\displaystyle \sec y^\circ[/tex]

Recall that tan(θ) = sin(θ) / cos(θ). Since sec(θ) = 1 / cos(θ), tan(θ) = sin(θ)sec(θ). Substitute:

[tex]\displaystyle \sin y^\circ \sec y^\circ = \frac{s}{t}[/tex]

Substitute:

[tex]\displaystyle \frac{s}{8}\sec y^\circ =\frac{s}{t}[/tex]

Solve for secant:

[tex]\displaystyle \sec y^\circ = \frac{8}{t}[/tex]

Hence, our answer is C.

Answer:

c. sec y° = 8/t

Step-by-step explanation:

I took the test

Answer:

A = 803.84

General Formulas and Concepts:

Symbols

Pre-Algebra

Order of Operations: BPEMDAS

Geometry

Area of a Circle Formula: A = πr²

Step-by-step explanation:

Step 1: Define

Identify

r = 16

Step 2: Find Area

Answer:

19 is the answer i think.

Answer:

you would receive 12.5% of the discount

Step-by-step explanation:

20% off of 62.50 is 50; so you would pay $50

the value of 4 tenths x hundreds.

4000

Answer:

40

Step-by-step explanation:

4/10 x 100 = 40

Answer:

37degrees

Step-by-step explanation:

In order to get the required angle, we will use the SOH, CAH, TOA identity.

Let;

Adjacent = 60m

Opposite = 45m

According to TOA:

tan theta = opp/adj

tan theta = 45/60

tan theta = 0.75

theta = arctan 0.75

theta =  36.86

Hence the  angle, to the  nearest degree, between her path and the longest side of the playground is 37degrees

Explanation:

In the LS column, you'll have these steps

3x - 1

3*4 - 1

12 - 1

11

Effectively, we replaced x with 4 and then simplified using PEMDAS.

And in the RS column, you'll have these steps

x + 7

4 + 7

11

We get the same thing at the bottom of each column. This shows that we end up with 11 = 11 after simplifying both sides. Therefore, we've confirmed that x = 4 is the solution to 3x - 1 = x + 7

Answer:

12 lemons

Step-by-step explanation:

CP of 15 lemon = ₹ 1

Profit = 25 %

SP of 15 lemons = [tex]\frac{100+profit}{100}*CP[/tex]

    [tex]= \frac{125}{100}*1\\\\= \frac{5}{4}[/tex]

SP of 15 lemons =  ₹  [tex]\frac{5}{4}[/tex]

Number of lemon for  ₹  [tex]\frac{5}{4}[/tex]  = 15

Number of lemons for  ₹  1 = 15 ÷ [tex]\frac{5}{4}[/tex]

                                            [tex]= 15 *\frac{4}{5}\\\\= 3*4\\\\= 12[/tex]

Answer:

answer to that is A 1.75 per pint

Respuesta:

A.) BA = 12 m

Explicación paso a paso:

Usando triángulos similares:

BA / AC = DE / EC

BA = x; AC = 20; DE = 3; EC = 5

POR ESO ; TENEMOS :

x / 20 = 3/5

Multiplicar en cruz:

5 * x = 20 * 3

5 veces = 60

5x / 5 = 60/5

x = 12

Answer:

-5x - 7y

Step-by-step explanation:

Answer:

q = -4   q=2

Step-by-step explanation:

q²+2q-8=0​

Factor

What 2 numbers multiply to -8 and add to 2

4*-2 = -8

4-2 = 2

(q+4)(q-2) =0

Using the zero product property

q+4 = 0  q-2 =0

q = -4   q=2

Answer:

50 km

Step-by-step explanation:

c = √a^2+b^2

=√30^2+40^2

= √900 + 1600

=√2500

= 50 km

Answer:

50km

Step-by-step explanation:

[tex]\boxed{\sf 60x^6y^4}[/tex]

Answer:

2688 cm³

Step-by-step explanation:

2058×704/539 = 2688 cm³

Answer:

[tex] \frac{(c - 7)}{2} = - 5 \\ c - 7 = - 5 \times 2 \\ c - 7 = - 10 \\ c = - 10 + 7 \\ c = - 3[/tex]

I hope I helped you^_^

Answer: x = 52

Step-by-step explanation:

You'd first deal with the 4 outside the parentheses, multiplying 10x-20 by 4. This would result in 40x - 80 = 200.

You'd then further isolate x, adding 80 on both sides, resulting in 40x = 2080. Divide by 40 on both sides and You'd get your answer of x = 52.

Answer:

..D).... x = 8..

Step-by-step explanation:

..x = 8..

Answer:

a) 3.5 cm / month

b y = 3.5(x-1) + 4, x >= 1

Step-by-step explanation:

the plant grew 32-4 = 28 cm in 9-1 = 8 months.

so, the growth rate is 28 cm / 8 months = 7 cm / 2 months, or (to norm it to a typical 1 unit denominator) 3.5 cm per month.

x = months of growth

y = height of plant

so, considering the "start" of 4 cm after one month, the equation is

y = 3.5(x-1) + 4, x >= 1

since we don't know anything about anything that happened before the first month was over, we cannot validate this equation for anything x<1.

Answer:

x = 15

Step-by-step explanation:

Using Pythagoras' identity in the right triangle

x² + 20² = 25²

x² + 400 = 625 ( subtract 400 from both sides )

x² = 225 ( take the square root of both sides )

x = [tex]\sqrt{225}[/tex] = 15

Answer:

Center ( 4 , -2) and r = 6

Step-by-step explanation:

x² - 8x + y² + 4y - 16 = 0

x² - 8x  + y² + 4 y = 16

x²- 2*4x + y² + 2 *2y = 16

(x² - 2*4x + 16 ) - 16 + (y² + 2*2y + 4) -4 = 16

(x - 4)² + (y + 2)² = 16 + 16 + 4

(x - 4)² + (y -[-2])² = 36

(x- 4)² + ( y - [-2] )² = 6²

Compare with (x - h)²+ (y - k)² = r²

Center ( 4 , -2) and r = 6

Answer:

[tex]a. \ \dfrac{625 \cdot m}{27 \cdot n^{11}}[/tex]

[tex]b. \ \dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}[/tex]

Step-by-step explanation:

The question relates with rules of indices

(a) The give expression is presented as follows;

[tex]\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}[/tex]

By expanding the expression, we get;

[tex]\dfrac{m^3 \times n^{-8} \times 5^4 \times m^4}{\left 3^3 \times m^6 \times n^3}[/tex]

Collecting like terms gives;

[tex]\dfrac{m^{(3 + 4 - 6)} \times 5^4}{ 3^3 \times n^{3 + 8}} = \dfrac{625 \cdot m}{27 \cdot n^{11}}[/tex]

[tex]\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}= \dfrac{625 \cdot m}{27 \cdot n^{11}}[/tex]

(b) The given expression is presented as follows;

[tex]x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \div (x \cdot y^n)^4[/tex]

Therefore, we get;

[tex]x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times x^{-4} \times y^{-4 \cdot n}[/tex]

Collecting like terms gives;

[tex]x^{3 \cdot m + 2 - 4} \times \left (y^{3 \cdot n - 3 -4 \cdot n}} \right ) = x^{3 \cdot m - 2} \times \left (y^{ - 3 -n}} \right ) = x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right )[/tex]

[tex]x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right ) = \dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}[/tex]

[tex]x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times x^{-4} \times y^{-4 \cdot n} =\dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}[/tex]

Answer:

y = 30

Step-by-step explanation:

Substitute x = 5 into the equation and solve for y

4(5) - y = - 10

20 - y = - 10 ( subtract 20 from both sides )

- y = - 30 ( multiply both sides by - 1 )

y = 30

Then (5, 30 ) is a solution of the equation

Answer:

( 5.5 , 4 )

Step-by-step explanation:

Use mid point formula shown in image

first x= (X+X)/2

x=(4+7)/2

x=11/2

x=5.5

y=(Y+Y)/2

y=(2+6)/2

y=8/2

y=4

(5.5,4)

Answer:

b the answer

Step-by-step explanation:

Answer:

[tex]\displaystyle x= \frac{y + 12}{3}[/tex]

General Formulas and Concepts:

Pre-Algebra

Equality Properties

Algebra I

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle 3x - y = 12[/tex]

Step 2: Solve for x

Answer:

x =  y + 12 / 3

Step-by-step explanation:

3x - y =12

3x = y + 12

x = y + 12 / 3