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3 years ago

If the car spends 2.5 hours going 35 miles per hour on the trip, how long does it spend going 35 miles per hours

Mathematics

1 answer:

Neporo4naja [7]3 years ago

6 0

its spends 2.5 hours going 35 miles per hour. so the answer is 2.5 because 35 miles as you said goes 2.5 hours

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Syrus is buying a tent with the dimensions shown below. The volume inside the tent is 363636 feet^3 3 cubed. Syrus isn't sure if

trasher [3.6K]

Answer:

The height of the tent = 3 feet

Step-by-step explanation:

Syrus is buying a tent with the dimensions shown below. The volume inside the tent is 36 feet^3. Syrus isn't sure if the tent will be tall enough for him to sit up inside. The tent is the shape of triangular prism whose length is 6 feet and width is 4 feet. What is the height of the tent?

Given:

Length of the tent = 6 feet

Width of the tent = 4 feet

Volume of the tent = 36

Solution:

Since the ten is in shape of triangular prism, so the volume of traingular prism is given as:

where represents length, represents width and represents height of the prism.

Plugging in the know values of the dimension of the tent and the volume to find the height of the tent.

Simplifying.

\frac{36}{12}=\frac{12h}{12}

3=h

∴ h=3

Thus, the height of the tent = 3 feet

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3 years ago

WILL MARK BRAINLIEST Triangle KON is dilated with a center at the origin and a

alexdok [17]

The answer is c: 0,6

3 0

3 years ago

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Find the volume of the cylinder. Use 3.14 for Pi. Round your answer to the nearest hundredth.the height is 13 and the radius is

NikAS [45]

Answer:

The volume of the cylinder is 2002 square units.

Step-by-step explanation:

It is given that,

Height of the cylinder, h = 13 units

Radius of the cylinder, r = 7 units

We need to find the volume of the cylinder. The formula of the volume of cylinder is given by :

$V=\pi r^2 h\\\\V=\dfrac{22}{7}\times 7^2 \times 13\\\\V=2002\ \text{unit}^2$

So, the volume of the cylinder is 2002 square units.

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3 years ago

‼️‼️‼️‼️‼️ is two correct? Also what’s number 3?‼️‼️‼️‼️

LenaWriter [7]

Answer:

C) P AND E

Step-by-step explanation:

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3 years ago

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Can I get help with finding the Fourier cosine series of F(x) = x - x^2

trapecia [35]

Assuming you want the cosine series expansion over an arbitrary symmetric interval $[-L,L]$ , $L\neq0$ , the cosine series is given by

$f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx$

You have

$a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx$
$a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}$
$a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)$
$a_0=-\dfrac{2L^2}3$

$a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx$

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

$\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}$

and so

$a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}$

So the cosine series for $f(x)$ periodic over an interval $[-L,L]$ is

$f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx$

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3 years ago