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2 years ago

5

Explain why Rolle's Theorem does not apply to the function even though there exist a and b such that f(a) = f(b). f(x) = cot x/

2 [π, 7π]

1 answer:

nydimaria [60]2 years ago

3 0

$cot(x/2)$ is not differentiable over the interval $[ \pi, 7\pi]$

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I think D is the answer.

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The base change property can help in what type of scenarios? HELP ME!!!!!

Mekhanik [1.2K]

<h3>Base Change Property</h3><h3 />

The Base Change Property is very helpful in scenarios related to simplifying equations where the logarithmic terms have a varying base.

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2 years ago

Estimate the product 5.25 × 8.89.

Anon25 [30]

Answer:

47.25

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1 year ago

Use the equation a = IaIâ

german

Answer:

a) $\:\:=\sqrt{14}\cdot \frac{\:\:}{\sqrt{14} }$

b) $\:\:=\sqrt{29} \cdot \frac{\:\:}{\sqrt{29} }$

c) $\:\:=7\cdot \frac{\:\:}{7}$

Step-by-step explanation:

a) Let <u>a</u>=<2,1,-3>

The magnitude of <u>a</u> is $|a|=\sqrt{2^2+1^2+(-3)^2}$

$|a|=\sqrt{4+1+9}=\sqrt{14}$

The unit vector in the direction of a is

$\hat{a}=\frac{\:\:}{\sqrt{14} }$

Using the relation $a=|a|\hat{a}$ , we have

$\:\:=\sqrt{14}\cdot \frac{\:\:}{\sqrt{14} }$

b) Let a=2i - 3j + 4k

$|a|=\sqrt{2^2+(-3)^2+4^2}$

$|a|=\sqrt{4+9+16}=\sqrt{29}$

$\hat{a}=\frac{\:\:}{\sqrt{29} }$

Using the relation $a=|a|\hat{a}$ , we have

$\:\:=\sqrt{29} \cdot \frac{\:\:}{\sqrt{29} }$

c) Let us first find the sum of <1, 2, -3> and <2, 4, 1> to get:

<1+2, 2+4, -3+1>=<3, 6, -2>

Let a=<3, 6, -2>

The magnitude is

$|a|=\sqrt{3^2+6^2+(-2)^2}$

$|a|=\sqrt{9+36+4}=\sqrt{49}=7$

The unit vector in the direction of <u>a</u> is

$\hat{a}=\frac{\:\:}{7}$

Using the relation $a=|a|\hat{a}$ , we have

$\:\:=7\cdot \frac{\:\:}{7}$

5 0

3 years ago