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3 years ago

Write equivalent expression to (5x - 3x2) - (5x - 2) . (10x + 7x2)

Mathematics

1 answer:

fiasKO [112]3 years ago

3 0

Man this is a hard question

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a 5 sided solid has the numbers 1,2,3,4, and 5. what is the probability of rolling two five-sided solids and getting a sum of ei

kipiarov [429]

So this is going to be alot of writing to show my thinking but ill bold the answer.

1,1
1,2
1,3
1,4
1,5

2,1
2,2
2,3
2,4
2,5

3,1
3,2
3,3
3,4
3,5

4,1
4,2
4,3
4,4
4,5

5,1
5,2
5,3
5,4
5,5

next ill mark all the ones that equal 4 or 8 when added together, with an x

1,1
1,2
x1,3
1,4
1,5

2,1
x2,2
2,3
2,4
2,5

x3,1
3,2
3,3
3,4
x3,5

4,1
4,2
4,3
x4,4
4,5

5,1
5,2
x5,3
5,4
5,5

that is 6 (that equal 4 or 8) out of 25

so your ratio would be 6:19

6 0

4 years ago

What is the slope of the line that passes through the points (4, 9) and (1, 6)?

Aleksandr [31]

a = ( 1 , 6 )

b = ( 4 , 9 )

$slope = \frac{y(b) - y(a)}{x(b) - x(a)} \\$

Now just need to put the coordinates in the above equation :

$slope = \frac{9 - 6}{4 - 1} \\$

$slope = \frac{3}{3} \\$

$slope = 1$

And we're done...♥️♥️♥️♥️♥️

5 0

3 years ago

Help please I beg u

MakcuM [25]

Answer:

56 i believe

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Using disks or washers, find the volume of the solid obtained by rotating the region bounded by the curves y=x−1 y=0 x=2 and x=5

Irina-Kira [14]

Using disks ...
$V= \pi \int\limits^5_2 {(x-1)^{2}} \, dx =\pi(\frac{1}{3}(5^{3}-2^{3})-(5^{2}-2^{2})+(5-2))=21\pi$

The volume is 21π units³ ≈ 65.97 units³

4 0

3 years ago

Find the area of the helicoid (or spiral ramp) with vector equation r(u, v) = ucos(v) i + usin(v) j + v k, 0 ≤ u ≤ 1, 0 ≤ v ≤ 9π

Natasha2012 [34]

Let $H$ denote the helicoid parameterized by

$\mathbb r(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+v\,\mathbf k$

for $0\le u\le1$ and $0\le v\le9\pi$ . The surface area is given by the surface integral,

$\displaystyle\iint_H\mathrm dS=\iint_H\|\mathbf r_u\times\mathbf r_v\|\,\mathrm du\,\mathrm dv$

We have

$\mathbf r_u=\dfrac{\partial\mathbf r(u,v)}{\partial u}=\cos v\,\mathbf i+\sin v\,\mathbf j$
$\mathbf r_v=\dfrac{\partial\mathbf r(u,v)}{\partial v}=-u\sin v\,\mathbf i+u\cos v\,\mathbf j+\mathbf k$
$\implies\mathbf r_u\times\mathbf r_v=\sin v\,\mathbf i-\cos v\,\mathbf j+u\,\mathbf k$
$\implies\|\mathbf r_u\times\mathbf r_v\|=\sqrt{1+u^2}$

So the area of $H$ is

$\displaystyle\iint_H\mathrm dS=\int_{v=0}^{v=9\pi}\int_{u=0}^{u=1}\sqrt{1+u^2}\,\mathrm du\,\mathrm dv$
$=\dfrac{9(\sqrt2+\sinh^{-1}(1))\pi}2$

5 0

3 years ago