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2 years ago

How are gallons and fluid ounces related

Mathematics

1 answer:

Leno4ka [110]2 years ago

6 0

They are both liquid measurements

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What is PR? A. 1 7/8 cm B. 10 cm C. 13 1/3 cm D. 15 cm

Bess [88]

Answer:

you answer should be b

Step-by-step explanation:

I am not 100% sure

8 0

2 years ago

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Hello, happy Friday, I am just here with some geometry questions.

klasskru [66]

Answer:

U'(-14, 8)

Step-by-step explanation:

The transformations include:

T<-2,2>·D3

This is a dilation by a scale factor of 3 and then translation 2 units left and 2 units up.

The transformation applied to the point U:

U(-4,2) → U'(-4*3 - 2, 2*3 + 2) = U'(-14, 8)

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2 years ago

In the diagram below, line AB and line CD intersect at point P. Name the angle that is always congruent to <APC

Alex Ar [27]

<dpb because they’re vertical angles (they’re directly across from each other)

5 0

3 years ago

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Match each interval with its corresponding average rate of change for q(x) = (x + 3)2. 1. -6 ≤ x ≤ -4 1 2. -3 ≤ x ≤ 0 -4 3. -6 ≤

MrMuchimi

The average rate of change of a function f(x) in an interval, a < x < b is given by
$\frac{f(b) - f(a)}{b - a}$

Given q(x) = (x + 3)^2

1.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -4 is given by $\frac{q(-4)-q(-6)}{-4-(-6)} = \frac{(-4+3)^2-(-6+3)^2}{-4+6} = \frac{1-9}{2} = \frac{-8}{2} =-4$

2.) The average rate of change of q(x) in the interval -3 ≤ x ≤ 0 is given by $\frac{q(0)-q(-3)}{0-(-3)} = \frac{(0+3)^2-(-3+3)^2}{0+3} = \frac{9-0}{3} = \frac{9}{3} =3$

3.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -3 is given by $\frac{q(-3)-q(-6)}{-3-(-6)} = \frac{(-3+3)^2-(-6+3)^2}{-3+6} = \frac{0-9}{3} = \frac{-9}{3} =-3$

4.) The average rate of change of q(x) in the interval -3 ≤ x ≤ -2 is given by $\frac{q(-2)-q(-3)}{-2-(-3)} = \frac{(-2+3)^2-(-3+3)^2}{-2+3} = \frac{1-0}{1} = \frac{1}{1} =1$

5.) The average rate of change of q(x) in the interval -4 ≤ x ≤ -3 is given by $\frac{q(-3)-q(-4)}{-3-(-4)} = \frac{(-3+3)^2-(-4+3)^2}{-3+4} = \frac{0-1}{1} = \frac{-1}{1} =-1$

6.) The average rate of change of q(x) in the interval -6 ≤ x ≤ 0 is given by $\frac{q(0)-q(-6)}{0-(-6)} = \frac{(0+3)^2-(-6+3)^2}{0+6} = \frac{9-9}{6} = \frac{0}{6} =0$

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2 years ago

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ANYONE PLEASE HELP!!!

VashaNatasha [74]

The answer is provided in the photo below:

8 0

2 years ago