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4 years ago

Given the table of the following quadratic function, which statement about this function is NOT true?

Mathematics

2 answers:

kodGreya [7K]4 years ago

7 0

C is the incorrect answer

kipiarov [429]4 years ago

6 0

Answer:

C. The minimum of the function f(x) is 8.

Step-by-step explanation:

The curve opens downward so 8 is a maximum.

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How do I solve -2i(-5+10i)

Snowcat [4.5K]

Answer:

-20i² + 10i

Step-by-step explanation:

- 2i (-5 + 10i)

distribute

- 2i (-5) + - 2i (10i)

10i - 20i²

- 20i² + 10i

4 0

3 years ago

7.29. hose fills a swimming pool at a rate of 0.5 gallons per second. If the pool's capacity is 9,000 gallons, then how many hou

Bond [772]

Answer:

It takes the hose 300 hours to fill the pool

Step-by-step explanation:

The hose can fill the pool at a rate of 0.5 gallons per second, which means that it can fill the pool at a rate of 1 gallon per 2 seconds. It needs to fill 9000 gallons of water, so assuming the ratio remains 1:2, it takes the hose 18000 (9000*2) seconds to fill the pool completely.

However, it is important to note that we have to answer in terms of hours. There are 60 seconds in an hour, so 18000 seconds is equal to 300 hours (18000/60).

7 0

2 years ago

Read 2 more answers

Plsss help I’ll give u brainlest and 10 points

Eddi Din [679]

The answer would be B I hope this helps.

5 0

3 years ago

A church window consisting of a rectangle topped by a semicircle is to have perimeter p . find the radius of the semicircle if t

Y_Kistochka [10]

Answer:

$r=\dfrac{p}{4+\pi}$

Step-by-step explanation:

Let r be the radius of the semicircle, then 2r is the width of the rectangle. Let y be the length of the rectangle. The perimeter of the window consists of two lengths, one width and length of semicircle, then

$p=2y+2r+\pi r.$

Express y:

$y=\dfrac{p-2r-\pi r}{2}.$

The area of the window is

$A=y\cdot 2r+\dfrac{1}{2}\pi r^2.$

Substitute y into the area expression:

$A=\dfrac{p-2r-\pi r}{2}\cdot 2r+\dfrac{1}{2}\pi r^2=pr-2r^2-\pi r^2+\dfrac{1}{2}\pi r^2=pr-2r^2-\dfrac{1}{2}\pi r^2.$

Find the derivative A':

$A'=p-4r-\pi r.$

Equate A' to 0:

$p-4r-\pi r=0\Rightarrow r=\dfrac{p}{4+\pi}.$

When $r$ then $A'>0$ and the function A is increasing, when $r>\dfrac{p}{4+\pi},$ then $A'$ and the function A is decreasing. This means that at point $r=\dfrac{p}{4+\pi}$ the function A takes it maximal value (the area is maximal).