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Anestetic [448]

3 years ago

15

At the beginning of the year Monica puts a set amount of money into her health benefit account. Every month she withdraws $15 fr

om this account for her contact lenses. After 3 months she has $255 left in her account.
Write an equation in slope-intercept form to represent the relationship between the time that has passed and the amount of money left in Monica's account. Let x represent the time in months and y represent the amount of money remaining in the account.

1 answer:

podryga [215]3 years ago

7 0

Answer:

An equation in the slope-intercept form will be:

y = 15x+210

Step-by-step explanation:

We know The slope-intercept form of the line equation

$y = mx+b$

where

m is the rate of change or slope

b is the y-intercept

Let x represent the time in months

Let y represent the amount of money remaining in the account.

Given that every month Monica withdraws $15 from this account for her contact lenses

Thus, the rate of change or slope = m = 15

Given that after 3 months she has $255 left in her account.

So, we take the point (3, 255)

substituting m = 15, x = 3 and y = 255 in the slope-intercept form to determine the y-intercept

y = mx+b

255 = 15(3) + b

switch sides

15(3) + b = 255

45 + b = 255

b = 255 - 45

b = 210

Now, substituting m = 15 and b = 210 in the slope-intercept form of the line equation

y = mx+b

y = 15x+210

Therefore, an equation in the slope-intercept form will be:

y = 15x+210

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A baby pool is shaped like a short, wide cylinder and holds 52 ft3 of water. A medium pool measures 4 times the height and width

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Answer:

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Step-by-step explanation:

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Required

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3 years ago

[30 POINTS] Please help!!!

Len [333]

Answer:

Part 1) $y=1.5x+5$

Part 2) $y=-(2/3)x-(11/3)$

Part 3) $y=0.25x+2.75$

Part 4) $y=-2x+5$

Part 5) $y=0.5x-1$

Part 6) The graph in the attached figure

Step-by-step explanation:

Part 1) we have

$m=3/2=1.5$

$point(-2,2)$

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

substitute

$y-2=1.5(x+2)$

$y=1.5x+3+2$

$y=1.5x+5$

Part 2) we know that

If two lines are perpendicular

then

the product of their slopes is equal to minus one

so

$m1*m2=-1$

the slope of the line 1 is equal to

$m1=1.5$

Find the slope m2

$1.5*m2=-1$

$m2=-2/3$

Find the equation of the line 2

we have

$m2=-2/3$

$point(-7,1)$

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

substitute

$y-1=(-2/3)(x+7)$

$y=-(2/3)x-(14/3)+1$

$y=-(2/3)x-(11/3)$

Part 3) we have

$m=1/4=0.25$

$point(1,3)$

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

substitute

$y-3=0.25(x-1)$

$y=0.25x-0.25+3$

$y=0.25x+2.75$

Part 4) we have

$m=-2$

$b=5$ -----> y-intercept

we know that

The equation of the line into slope intercept form is equal to

$y=mx+b$

substitute the values

$y=-2x+5$

Part 5) we have that

The slope of the line 4 is equal to $-2$

so

the slope of the line perpendicular to the line 4 is equal to

$-2*m=-1\\m=(1/2)=0.5$

therefore

in this problem we have

$m=0.5$

$point(-2,-2)$

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

substitute

$y+2=0.5(x+2)$

$y=0.5x+1-2$

$y=0.5x-1$

Part 6)

using a graphing tool

see the attached figure

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3 years ago