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3 years ago

Using the slope of 7 and y-intercept of (0, 16), write an equation of the line

Mathematics

1 answer:

Katen [24]3 years ago

6 0

Answer:

y = 7x + 16

Step-by-step explanation:

Slope (m) = 7

y - intercept (b) = 16..{from (0, 16)}

Equation of line in slope intercept form is given as:

$y = mx + b \\ \therefore \: y = 7 \times x + 16 \\ \huge \red{ \boxed{ \therefore \: y = 7 x + 16 }}$

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Plot the vertices (-2, -8), (-2, 2), and (8, 2).

charle [14.2K]

Answer:

(8, -8)
100 square units

Step-by-step explanation:

The diagonals of a rectangle have the same midpoint, so for points A, B, C, D, we must have ...

(A+C)/2 = (B+D)/2

D = A + C - B

D = (-2, -8) +(8, 2) -(-2, 2) = (-2+8+2, -8+2-2)

D = (8, -8)

The side lengths are 2-(-8) = 10, and 8-(-2) = 10. The area is the product of the side lengths, so is 10×10 = 100 square units.

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<em>Comment on side lengths</em>

When the end points of one side are on the same vertical line, the length of that side is the difference of y-coordinates. When the points lie on the same horizontal line, the side length is the difference of x-coordinates.

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3 years ago

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Trava [24]

Answer:

5n units

Step-by-step explanation:

C=2πr =2·π·2.5

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2 years ago

What do a parallelogram and kite have in common?

grigory [225]

They both have 4 congruent sides and angles.

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3 years ago

3x - 6 + 4 = -20

V125BC [204]

Answer:

I'm not sure what all that extra stuff is supposed to mean, but the answer to the top question is: x = -6

Step-by-step explanation:

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2 years ago

The fish population of Lake Collins is decreasing at a rate of 5% per year. In 2004 there were about 1,350 fish. Write an expone

ad-work [718]

Answer:

Part 1) The exponential function is equal to $y=1,350(0.95)^{x}$

Part 2) The population in 2010 was $992\ fish$

Step-by-step explanation:

Part 1) Write an exponential decay function that models this situation

we know that

In this problem we have a exponential function of the form

$y=a(b)^{x}$

where

y ----> the fish population of Lake Collins since 2004

x ----> the time in years

a is the initial value

b is the base

we have

$a=1,350\ fish$

$b=(100\%-5\%)=95\%=0.95$

substitute

$y=1,350(0.95)^{x}$ ----> exponential function that represent this scenario

Part 2) Find the population in 2010

we have

$y=1,350(0.95)^{x}$

For $x=(2010-2004)=6\ years$

substitute

$y=1,350(0.95)^{6}=992\ fish$

7 0

3 years ago