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2 years ago

Please help! thank you!

Mathematics

1 answer:

anyanavicka [17]2 years ago

7 0

Answer:

B. r=7.5

yan po ang correct answer

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3h = (2p – 1) + h Please someone help

Natalka [10]

Are you solving p or h
This is the answer for h h=p-1/2
This is the answer for p p=h+1/2

4 0

2 years ago

A rectangle park measures 300 ft by 400 ft. A sidewalk runs diagonally from one comer to the opposite corner. Find the length o

DochEvi [55]

The length of side walk is 500 feet

Solution:

Given that, A rectangle park measures 300 ft by 400 ft

Length = 300 feet

Width = 400 feet

A sidewalk runs diagonally from one comer to the opposite corner

We have to find the length of side walk

Which means, we have to find the length of diagonal of rectangle

The diagonal of rectangle is given by formula:

$d = \sqrt{w^2+l^2}$

Where,

d is the length of diagonal

w is the width and l is the length of rectangle

Substituting the values in formula, we get

$d = \sqrt{400^2+300^2}\\\\d = \sqrt{160000+90000}\\\\d = \sqrt{250000}\\\\d = 500$

Thus length of side walk is 500 feet

5 0

3 years ago

What is the equation for table A. What is the equation for table B.

xeze [42]

Answer:

A: y=-2/3x + 3

B: y=x-2

solution: (3,1)

Step-by-step explanation:

7 0

2 years ago

Suppose that an experiment is repeated four times. A certain event has probability 1/10 in a single repetition of the

frez [133]

Answer:

2/5

Step-by-step explanation:

1/10 times 4

4/10 equals to 2/5

5 0

2 years 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

Please help! thank you!​

Please help! thank you!