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mr Goodwill [35]
2 years ago
6

How to find each angle indicated

Mathematics
2 answers:
Natalija [7]2 years ago
8 0
The answer is 80 because all triangles equal 180
Basile [38]2 years ago
5 0
The Answer Going To Be 80
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Can somebody help with this problem
Semmy [17]
Well, if every box is a unit, it would be 6 units in length.
4 0
2 years ago
What is 1000000000000000000000000 called?
STALIN [3.7K]

Answer:

septillion

Step-by-step explanation:

3 0
2 years ago
To which real number sets does the number 13 belong?
Crazy boy [7]
13 is a whole number, integer, natural number and rational number. Rational number because it can be written as a fraction like 13/1. It is a whole number because it doesn't have any decimals. It is an integer because an integer is a positive or negative whole number and it is a natural number because a natural number is any positive number greater than 0.
3 0
3 years ago
Which one of the quotients is correct? A) 876 ÷ 43 = 20r17B) 876 ÷ 43 = 20r16C) 876 ÷ 43 = 20r7D) 876 ÷ 43 = 20r6
Olegator [25]

To check which of the quotients is correct, multiply 43 times 20 and add the remainder. The result must be equal to 876.

First, notice that 20 times 43 equals 860.

A)

The remainder is 17. 860 + 17 = 877, which is not equal to 876.

B)

The remainder is 16. 860 + 16 = 876, which is equal to 876.

C)

The remander is 7. 860 + 7 = 867, which is not equal to 876.

D)

The remainder is 6. 860 + 6 = 866, which is not equal to 876.

Since 20*43 + 16 = 876, then the correct quotient is shown in option B:

876 ÷ 43 = 20 r 16

4 0
1 year ago
Consider the function ​f(x)equalscosine left parenthesis x squared right parenthesis. a. Differentiate the Taylor series about 0
dybincka [34]

I suppose you mean

f(x)=\cos(x^2)

Recall that

\cos x=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}

which converges everywhere. Then by substitution,

\cos(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{(x^2)^{2n}}{(2n)!}=\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n)!}

which also converges everywhere (and we can confirm this via the ratio test, for instance).

a. Differentiating the Taylor series gives

f'(x)=\displaystyle4\sum_{n=1}^\infty(-1)^n\frac{nx^{4n-1}}{(2n)!}

(starting at n=1 because the summand is 0 when n=0)

b. Naturally, the differentiated series represents

f'(x)=-2x\sin(x^2)

To see this, recalling the series for \sin x, we know

\sin(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^{n-1}\frac{x^{4n+2}}{(2n+1)!}

Multiplying by -2x gives

-x\sin(x^2)=\displaystyle2x\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n+1)!}

and from here,

-2x\sin(x^2)=\displaystyle 2x\sum_{n=0}^\infty(-1)^n\frac{2nx^{4n}}{(2n)(2n+1)!}

-2x\sin(x^2)=\displaystyle 4x\sum_{n=0}^\infty(-1)^n\frac{nx^{4n}}{(2n)!}=f'(x)

c. This series also converges everywhere. By the ratio test, the series converges if

\displaystyle\lim_{n\to\infty}\left|\frac{(-1)^{n+1}\frac{(n+1)x^{4(n+1)}}{(2(n+1))!}}{(-1)^n\frac{nx^{4n}}{(2n)!}}\right|=|x|\lim_{n\to\infty}\frac{\frac{n+1}{(2n+2)!}}{\frac n{(2n)!}}=|x|\lim_{n\to\infty}\frac{n+1}{n(2n+2)(2n+1)}

The limit is 0, so any choice of x satisfies the convergence condition.

3 0
3 years ago
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