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3 years ago

For a convex 41-gon, what is the sum of the measures of the interior angles?

1 answer:

7 0

Given:

convex 41-gon
n = 41

Required:

Sum of interior angles.

Solution:

To solve for the sum of interior angles, we can use the formula:

Sum of interior angles = 180(n-2)°

Substituting,

Sum of interior angles = 180(41-2)°

Sum of interior angles = 7020°

Therefore, the sum of the measures of the interior angle of a convex 41-gon is 7020°.

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You rent an apartment that costs $900 per month during the first year, but the rent is set to go up $170 per year. What would be

vichka [17]

The monthly rent of the apartment in the 10th year is $1041.67.

<h3>What is the monthly rent during the 10th year of living in the apartment?</h3>

The first step is to determine the yearly rent in the first year of living in the apartment.

Yearly rent = $900 x 12 = $10,800

The second step is to determine the yearly rent in the 10th year

$10,800 + (170 x 10) = $12,500

The third step is to determine the monthly rent

Monthly rent = $12500 / 12 = $1041.67

To learn more about division, please check: brainly.com/question/13281206

3 0

2 years ago

EXAMPLE 2 Prove that 9ex is equal to the sum of its Maclaurin series. SOLUTION If f(x) = 9ex, then f (n + 1)(x) = for all n. If

amm1812

Answer:

To Prove: $9e^x$ is equal to the sum of its Maclaurin series.

Step-by-step explanation:

If $f(x) = 9e^x$ , then $f ^{(n + 1)(x)} =9e^x$ for all n. If d is any positive number and |x| ≤ d, then $|f^{(n + 1)(x)}| = 9e^x\leq 9e^d.$

So Taylor's Inequality, with a = 0 and M = $9e^d$ , says that $|R_n(x)| \leq \dfrac{9e^d}{(n+1)!} |x|^{n + 1} \:for\: |x| \leq d.$

Notice that the same constant $M = 9e^d$ works for every value of n.

But, since $lim_{n\to\infty}\dfrac{x^n}{n!} =0 $ for every real number x$$ ,

We have $lim_{n\to\infty} \dfrac{9e^d}{(n+1)!} |x|^{n + 1} =9e^d lim_{n\to\infty} \dfrac{|x|^{n + 1}}{(n+1)!} =0$

It follows from the Squeeze Theorem that $lim_{n\to\infty} |R_n(x)|=0$ and therefore $lim_{n\to\infty} R_n(x)=0$ for all values of x.

$THEOREM\\If f(x)=T_n(x)+R_n(x), $where $T_n $is the nth degree Taylor Polynomial of f at a and $ lim_{n\to\infty} R_n(x)=0 \: for \: |x-a|$

By this theorem above, $9e^x$ is equal to the sum of its Maclaurin series, that is,

$9e^x=\sum_{n=0}^{\infty}\frac{9x^n}{n!}$ for all x.

6 0

3 years ago

5 7/5-5 2/5-3 4/5 equal what

Katyanochek1 [597]

Answer:

-78.424

Step-by-step explanation:

8 0

3 years ago

Martina creates the graph of function g by applying a transformation to function f

dsp73

Answer:

Step-by-step explanation:

f(x)=4^x-2

g(x)=4^x+7

it is horizontal shift 9 unit to the left

hope it works

3 0

3 years ago

Can someone help me with this problem

Alex_Xolod [135]

Answer:

Step-by-step explanation:

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

A(-3, 0) , B(3, 2)

$AB=\sqrt{(3-(-3))^2+(2-0)^2}=\sqrt{6^2+2^2}=\sqrt{36+4}=\sqrt{40}=\sqrt{4\cdot10}=\sqrt4\cdot\sqrt{10}=2\sqrt{10}$

A(-3, 0), D(-2, -3)

$AD=\sqrt{(-2-(-3))^2+(-3-0)^2}=\sqrt{1^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}$

AB = CD and AD = BC

The area of a rectangle:

$A=(AB)(AD)$

Substitute:

$A=(2\sqrt{10})(\sqrt{10})=(2)(10)=20$

The perimeter of a rectangle:

$P=2(AB+AD)$

Substitute:

$P=2(2\sqrt{10}+\sqrt{10})=2(3\sqrt{10})=6\sqrt{10}$

4 0

3 years ago