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3 years ago

Please help with these questions for alot of points and brainiest

Mathematics

2 answers:

Sliva [168]3 years ago

6 0

5. To figure this out you must find when y = 0

-16t^2 = -200

t^2 = 12.5

Square root

t = 3.54 C

Axis of symetry = -.5 (always a vertical line down vertex)

Vertex = (-.5, -6.5) (the minimum of the graph)

7. To find maximum height you must find the vertex

The maximum height is 549 feet after 5.75 seconds

9. Solve for X

x^2 = 1

Square root

X = plus or minus 1

Hitman42 [59]3 years ago

4 0

Answer:

5. To figure this out you must find when y = 0

-16t^2 = -200

t^2 = 12.5

Square root

t = 3.54 C

Axis of symetry = -.5 (always a vertical line down vertex)

Vertex = (-.5, -6.5) (the minimum of the graph)

7. To find maximum height you must find the vertex

The maximum height is 549 feet afte

Read more on Brainly.com - brainly.com/question/11635126#readmore

Step-by-step explanation:

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Steven has a stamp collection with 32 stamps. Only 1/4 of the stamps are from the United States. How many stamps are from the U.

Natasha_Volkova [10]

Answer:

I believe the answer would be 8 because if you divide 32 by 4 you get eight and that would be 1/4 of 32

Step-by-step explanation:

5 0

3 years ago

In rectangle ABCD, AC = -x + 21, BP = -9x + 19, and DP = -y + 5. Find the value of y.

creativ13 [48]

Answer:

y=-5

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

The diagonals of a rectangle are equal and bisect each other

$BP=\frac{1}{2}AC$

$DP=BP$

step 1

Find the value of x

$BP=\frac{1}{2}AC$

substitute the given values

$(-9x+19)=\frac{1}{2} (-x+21)$

solve for x

Multiply by 2 both sides

$-18x+38=-x+21$

$18x-x=38-21\\17x=17\\x=1$

step 2

Find the value of y

$DP=BP$

we have

$BP=-9x+19$

$DP=-y+5$

substitute

$-y+5=-9x+19$

substitute the value of x

$-y+5=-9(1)+19\\-y+5=10\\y=5-10\\y=-5$

5 0

3 years ago

-2 < n s6<br><br> n is an integer.<br><br> What are the possible values of n?

alex41 [277]

This is what I found

4 0

3 years ago

Solve using Fourier series.

Olin [163]

With $2L=\pi$ , the Fourier series expansion of $f(x)$ is

$\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos\dfrac{n\pi x}L+\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L$
$\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos2nx+\sum_{n\ge1}b_n\sin2nx$

where the coefficients are obtained by computing

$\displaystyle a_0=\frac1L\int_0^{2L}f(x)\,\mathrm dx$
$\displaystyle a_0=\frac2\pi\int_0^\pi f(x)\,\mathrm dx$

$\displaystyle a_n=\frac1L\int_0^{2L}f(x)\cos\dfrac{n\pi x}L\,\mathrm dx$
$\displaystyle a_n=\frac2\pi\int_0^\pi f(x)\cos2nx\,\mathrm dx$

$\displaystyle b_n=\frac1L\int_0^{2L}f(x)\sin\dfrac{n\pi x}L\,\mathrm dx$
$\displaystyle b_n=\frac2\pi\int_0^\pi f(x)\sin2nx\,\mathrm dx$

You should end up with

$a_0=0$
$a_n=0$
(both due to the fact that $f(x)$ is odd)
$b_n=\dfrac1{3n}\left(2-\cos\dfrac{2n\pi}3-\cos\dfrac{4n\pi}3\right)$

Now the problem is that this expansion does not match the given one. As a matter of fact, since $f(x)$ is odd, there is no cosine series. So I'm starting to think this question is missing some initial details.

One possibility is that you're actually supposed to use the even extension of $f(x)$ , which is to say we're actually considering the function

$\varphi(x)=\begin{cases}\frac\pi3&\text{for }|x|\le\frac\pi3\\0&\text{for }\frac\pi3$

and enforcing a period of $2L=2\pi$ . Now, you should find that

$\varphi(x)\sim\dfrac2{\sqrt3}\left(\cos x-\dfrac{\cos5x}5+\dfrac{\cos7x}7-\dfrac{\cos11x}{11}+\cdots\right)$

The value of the sum can then be verified by choosing $x=0$ , which gives

$\varphi(0)=\dfrac\pi3=\dfrac2{\sqrt3}\left(1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots\right)$
$\implies\dfrac\pi{2\sqrt3}=1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots$

as required.

5 0

3 years ago

Please Help me solve this (Who ever answers correctly will get brainlest)

o-na [289]

Answer:

1.52

……………………………………………………

5 0

2 years ago

Read 2 more answers