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2 years ago

Using the Pythagorean method solve the problem and find the area of the triangle

Mathematics

1 answer:

denis-greek [22]2 years ago

8 0

Answer:pythagoream theorem applies to right angle triangle.

Formula

C^2= a^2 + b^ 2

Where a&b are the two sides of the triangle so we need to find side c.

C^2 = 35.42^2+ 21.64^2

C^2 = 1254.57 + 468.29

C^2 = 1722.86

C = square root of 1722.86

C= 41.51m

Area of triangle

A = ab/2

A = 35.42 x 21.64/2

A = 383.24.

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

The constant is 4.

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2 years ago

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masya89 [10]

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2 years ago

One ticket costs $15.45. Three tickets would cost?

Troyanec [42]

46.35 is the answer

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2 years ago

Evaluate the given question

Natali5045456 [20]

The value of the given expression is 1/16

<h3>Exponential functions</h3>

Given the exponential function below as shown;

(1/2)^4

This means the product of 1/2 in 4 places. This can be expressed as;

(1/2)^4 = 1/2 * 1/2 * 1/2 *1/2

(1/2)^4 = 1/4 * 1/4

(1/2)^4 =1/16

Hence the value of the given expression is 1/16

Learn more on exponent here: brainly.com/question/725335

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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