Given g(x) how do I find the 4 right endpoint rectangles

Please explain I don’t I understand help

djyliett [7]

Answer:

Step-by-step explanation:

f (x) = (5(-4)) + a) (3/5 + 4) ( 6 + b)

f (x) = (-20 + a) (6 2/5) (6 + b)

f ( -20 ) = -20 + -20 (a) (6 2/5) (6 + b)

f ( -20 - 6 2/5 ) = (a) (6 2/5 -6 2/5) (6 + b)

f (-26 2/5 - 6 ) = (a) (6-6 +b)

f (-32 2/5-6) = (a) (b)

4 0

2 years ago

ASAP ANSWER PLZ THANK YOU!

SVETLANKA909090 [29]

Answer:

The last one

Step-by-step explanation:

The last table represents a linear function, y = -2x + 3

3 0

3 years ago

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Ann is adding water to a swimming pool at a constant rate. The table below shows the amount of water in the pool after different

N76 [4]

Answer:

(a) As time increases, the amount of water in the pool increases.

11 gallons per minute

(b) 65 gallons

Step-by-step explanation:

From inspection of the table, we can see that as time increases, the amount of water in the pool increases.

We are told that Ann adds water at a constant rate. Therefore, this can be modeled as a linear function.

The rate at which the water is increasing is the rate of change (which is also the slope of a linear function).

Choose 2 ordered pairs from the table:

$\textsf{let}\:(x_1,y_1)=(8, 153)$

$\textsf{let}\:(x_2,y_2)=(12,197)$

Input these into the slope formula:

$\textsf{slope}\:(m)=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{197-153}{12-8}=\dfrac{44}{4}=11$

Therefore, the rate at which the water in the pool is increasing is:

11 gallons per minute

To find the amount of water that was already in the pool when Ann started adding water, we need to create a linear equation using the found slope and one of the ordered pairs with the point-slope formula:

$y-y_1=m(x-x_1)$

$\implies y-153=11(x-8)$

$\implies y-153=11x-88$

$\implies y=11x-88+153$

$\implies y=11x+65$

When Ann had added no water, x = 0. Therefore,

$y=11(0)+65$

$y=65$

So there was 65 gallons of water in the pool before Ann starting adding water.

3 0

2 years ago

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The length of each side of an equilateral triangle is increased by 20%, resulting in triangle ABC. If the length of each side of

kompoz [17]

Answer: Area of ΔABC is 2.25x the area of ΔDEF.

Step-by-step explanation: Because equilateral triangle has 3 equal sides, area is calculated as

$A=\frac{\sqrt{3} }{4} a^{2}$

with a as side of the triangle.

Triangle ABC is 20% bigger than the original, which means its side (a₁) measures, compared to the original:

a₁ = 1.2a

Then, its area is

$A_{1}=\frac{\sqrt{3} }{4}(1.2a)^{2}$

$A_{1}=\frac{\sqrt{3} }{4}1.44a^{2}$

Triangle DEF is 20% smaller than the original, which means its side is:

a₂ = 0.8a

So, area is

$A_{2}=\frac{\sqrt{3} }{4} (0.8a)^{2}$

$A_{2}=\frac{\sqrt{3} }{4} 0.64a^{2}$

Now, comparing areas:

$\frac{A_{1}}{A_{2}}= (\frac{\sqrt{3}.1.44a^{2} }{4})(\frac{4}{\sqrt{3}.0.64a^{2} } )$

$\frac{A_{1}}{A_{2}} =$ 2.25

The area of ΔABC is 2.25x greater than the area of ΔDEF.

7 0

2 years ago

Where the numerator is larger than the denominator

gregori [183]

A fraction with a numerator that is greater than or equal to the denominator is known as an improper fraction.

4 0

3 years ago

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