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3 years ago

9

Samantha's annual college tuition is $18,541.89. The college offers a monthly payment plan for 9 equal payments. How much would

each monthly payment be?

1 answer:

Andru [333]3 years ago

4 0

Answer:

2,060.21

Step-by-step explanation:

:)

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9. If f(x) = 5x-4 and f(2a) =156 , then what is the value of a?

solmaris [256]

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

$f(x) = 5x - 4$

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

$f(2a) = 5(2a) - 4$

$f(2a) = 10a - 4$

$156 = 10a - 4$

$10a - 4 = 156$

Add sides 4

$10a - 4 + 4 = 156 + 4$

$10a = 160$

Divide sides by 10

$\frac{10a}{10} = \frac{160}{10} \\$

$a = 16$

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

7 0

3 years ago

A bag contains 2 white marbles and 7 purple marbles. Two marbles are drawn at random. One marble is drawn and not replaced, then

skelet666 [1.2K]

A. likely

B. unlikely

C. There is a greater chance in picking to purple marbles because there are less of the white and more purple.

Hallie should spin the wheel because u have a greater chance of landing on a 5 then rolling two number cubes and hoping both number cubes come to equal 5.

4 0

3 years ago

Read 2 more answers

Triangle A″B″C″ is formed using the translation (x + 1, y + 1) and the dilation by a scale factor of 3 from the origin. Which eq

FromTheMoon [43]

Answer:

$B''C''=3BC$

Step-by-step explanation:

we know that

The length of segment B''C'' is equal to the length of segment BC multiplied by the scale factor

step 1

Find the length of segment BC

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have the points

B(1,-3),C(-3,-3)

substitute the values

$BC=\sqrt{(-3+3)^{2}+(-3-1)^{2}}$

$BC=\sqrt{(0)^{2}+(-4)^{2}}$

$BC=4\ units$

The segment B''C'' is equal to the segment BC multiplied by the scale factor

The scale factor is 3

so

$B''C''=BC(3)$

$B''C''=3BC$

substitute

$B''C''=3(4)=12\ units$

4 0

3 years ago

Read 2 more answers

If f(x) = 3x + 7, which of these is the inverse of f(x)?

dolphi86 [110]

Answer:

f-1(x)=x-7/3

Step-by-step explanation:

7 0

3 years ago

A 1.5-mm layer of paint is applied to one side of the following surface. Find the approximate volume of paint needed. Assume tha

Mariulka [41]

Answer:

V = 63π / 200 m^3

Step-by-step explanation:

Given:

- The function y = f(x) is revolved around the x-axis over the interval [1,6] to form a spherical surface:

y = √(42*x - x^2)

- The surface is coated with paint with uniform layer thickness t = 1.5 mm

Find:

The volume of paint needed

Solution:

- Let f be a non-negative function with a continuous first derivative on the interval [1,6]. The Area of surface generated when y = f(x) is revolved around x-axis over the interval [1,6] is:

$S = 2*\pi \int\limits^a_b { [f(x)*\sqrt{1 + f'(x)^2} }] \, dx$

- The derivative of the function f'(x) is as follows:

$f'(x) = \frac{21-x}{\sqrt{42x-x^2} }$

- The square of derivative of f(x) is:

$f'(x)^2 = \frac{(21-x)^2}{42x-x^2 }$

- Now use the surface area formula:

$S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2} *\sqrt{1 + \frac{(21-x)^2}{42x-x^2 } }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+(21-x)^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+441-42x+x^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{441} }] \, dx\\S = 2*\pi \int\limits^6_1 { 21} \, dx\\\\S = 42*\pi \int\limits^6_1 { dx} \,\\\\S = 42*\pi [ 6 - 1 ]\\\\S = 42*5*\pi \\\\S = 210\pi$

- The Volume of the pain coating is:

V = S*t

V = 210*π*3/2000

V = 63π / 200 m^3

8 0

3 years ago

Read 2 more answers