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Galina-37 [17]
2 years ago
5

What is 26.5 rounded to the nearest hundreds

Mathematics
1 answer:
scZoUnD [109]2 years ago
7 0
5 rounds up 6 rounds up 2 stays the same what is the answer?
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Give examples of numbers that you would multiply by 3 to get an answer that is greater than 3, equal to 3, or less than 3
oksian1 [2.3K]

Answer:

1. 3 x <u>3</u> = 9

2. 3 x <u>1</u> =3

3. 3 x <u>0.4</u> = 1.2

Step-by-step explanation: Hope this helps :)

6 0
2 years ago
The angle of depression from the basketball hoop to Claire’s eyes is 9.2°. The vertical distance from the ground to Claire’s eye
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8 0
3 years ago
Consider the following function. f(x) = 16 − x2/3 Find f(−64) and f(64). f(−64) = f(64) = Find all values c in (−64, 64) such th
VARVARA [1.3K]

Answer:

This does not contradict Rolle's Theorem, since f '(0) = 0, and 0 is in the interval (−64, 64).

Step-by-step explanation:

The given function is

f(x)=16-\frac{x^2}{3}

To find f(-64), we substitute x=-64 into the function.

f(-64)=16-\frac{(-64)^2}{3}

f(-64)=16-\frac{4096}{3}

f(-64)=-\frac{4048}{3}

To find f(64), we substitute x=64 into the function.

f(64)=16-\frac{(64)^2}{3}

f(64)=16-\frac{4096}{3}

f(64)=-\frac{4048}{3}

To find f'(c), we must first find f'(x).

f'(x)=-\frac{2x}{3}

This implies that;

f'(c)=-\frac{2c}{3}

f'(c)=0

\Rightarrow -\frac{2c}{3}=0

\Rightarrow -\frac{2c}{3}\times -\frac{3}{2}=0\times -\frac{3}{2}

c=0

For this function to satisfy the Rolle's Theorem;

It must be continuous on [-64,64].

It must be differentiable  on (-64,64).

and

f(-64)=f(64).

All the hypotheses are met, hence this does not contradict Rolle's Theorem, since f '(0) = 0, and 0 is in the interval (−64, 64) is the correct choice.

6 0
2 years ago
Read 2 more answers
What is the value of f(x) when x = 3?
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It's not specific enough to solve...so far it would only be f(3)
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