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3 years ago

10

What is 5mm as a radius plz help

1 answer:

exis [7]3 years ago

6 0

Well, assuming 5mm is the diameter, then 2.5 mm is the radius.

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The graph below belongs to which function family?

svet-max [94.6K]

The answer is the last one (absolute value)

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3 years ago

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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.

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3 years ago

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Which answer correctly estimates the quotient of 102.02 ÷ 4.011?

BigorU [14]

Well the answer is A) about 25

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3 years ago

Find the x-intercept of the parabola with vertex (1,-13) and y-intercept (0,-11).

Advocard [28]

1. find equation

general solutions are:
$y = a x^{2} +bx +c \\ \\ \frac{dy}{dx} = 2ax +b$

for P(0,-11)

$-11 = 0a + 0b +c$ ⇒ $c = -11$

for p(1,-13)
$-13 = 1a + 1b -11 \\ a+b = -2$

and

$0 = 2a + b \\ b = -2a$

solve for a and b:
$a - 2a = -2 \\ a = 2 \\ b = -4$

the total equation is now:

$y = 2 x^{2} -4x -11$

To find the x-intercept set y=0 and solve for x

$0 = 2 x^{2} -4x-11$

4 0

3 years ago

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Tan9°. tan27°. tan45°. tan72°. tan81°= 1<br><br>could u plz solve it with clear process

Inga [223]

<u>Correct</u><u> </u><u>question</u><u>:</u><u>-</u>

<u>Prove </u><u>that </u><u>tan9.</u><u>t</u><u>a</u><u>n</u><u>1</u><u>7</u><u>.</u><u>t</u><u>a</u><u>n</u><u>4</u><u>5</u><u>.</u><u>t</u><u>a</u><u>n</u><u>7</u><u>3</u><u>.</u><u>t</u><u>a</u><u>n</u><u>8</u><u>1</u><u>=</u><u>1</u>

<u>LHS</u>

$\\ \sf\longmapsto tan9.tan17.tan45.tan73.tan81$

$\\ \sf\longmapsto tan9.tan81.tan17.tan73.tan45$

$\\ \sf\longmapsto tan(90-81).tan81.tan(90-73).tan73.1$

$\\ \sf\longmapsto cot81.tan81.cot73.tan73$

$\\ \sf\longmapsto 1.1$

$\\ \sf\longmapsto 1$

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2 years ago

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