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3 years ago

9x-1=11 how would I solvr this problem

1 answer:

3 0

Add one to both side so its 9x =12 then divide 9 on both sides to get x alone and you have 9/12 simplified 3/4 in decimal form .75

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I need help on how to do combinations

Veronika [31]

Ok sure! Let's start with the marbles first. We have one red one blue and one green. There is a red and a blue cube. All we do is list the possible combinations of one marble and one cube. The answer would be : 1 blue marble, 1 blue cube. 1 red marble, 1 blue cube. 1 green marble, 1 blue cube. Then again but with the red cube. So in all, there are 6 possible combinations. Hope this helps. ☺️

3 0

3 years ago

Evaluate the surface integral:S

rjkz [21]

Assuming $S$ does not include the plane $z=0$ , we can parameterize the region in spherical coordinates using

$\mathbf r(u,v)=\left\langle3\cos u\sin v,3\sin u\sin v,3\cos v\right\rangle$

where $0\le u\le2\pi$ and $0\le v\le\dfrac\pi/2$ . We then have

$x^2+y^2=9\cos^2u\sin^2v+9\sin^2u\sin^2v=9\sin^2v$
$(x^2+y^2)=9\sin^2v(3\cos v)=27\sin^2v\cos v$

Then the surface integral is equivalent to

$\displaystyle\iint_S(x^2+y^2)z\,\mathrm dS=27\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^2v\cos v\left\|\frac{\partial\mathbf r(u,v)}{\partial u}\times \frac{\partial\mathbf r(u,v)}{\partial u}\right\|\,\mathrm dv\,\mathrm du$

We have

$\dfrac{\partial\mathbf r(u,v)}{\partial u}=\langle-3\sin u\sin v,3\cos u\sin v,0\rangle$
$\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle3\cos u\cos v,3\sin u\cos v,-3\sin v\rangle$
$\implies\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle-9\cos u\sin^2v,-9\sin u\sin^2v,-9\cos v\sin v\rangle$
$\implies\left\|\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}\|=9\sin v$

So the surface integral is equivalent to

$\displaystyle243\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv\,\mathrm du$
$=\displaystyle486\pi\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv$
$=\displaystyle486\pi\int_{w=0}^{w=1}w^3\,\mathrm dw$

where $w=\sin v\implies\mathrm dw=\cos v\,\mathrm dv$ .

$=\dfrac{243}2\pi w^4\bigg|_{w=0}^{w=1}$
$=\dfrac{243}2\pi$

4 0

3 years ago

Using the formula T=7V+W, find the value of V when T = 26 and W = 5

ivann1987 [24]

26=7v + 5
21=7v
V=21/3
V=3

5 0

3 years ago

Read 2 more answers

Find the sum of first 15 terms of an AP whose 4th and 9th t terms are -15 and -30 respectively.

Serga [27]

Answer:

Sum of the first 15 terms = -405

Step-by-step explanation:

a + 3d = -15 (1)

a + 8d = -30 (2)

Where,

a = first term

d = common difference

n = number of terms

Subtract (1) from (1)

8d - 3d = -30 - (-15)

5d = -30 + 15

5d = -15

d = -15/5

= -3

d = -3

Substitute d = -3 into (1)

a + 3d = -15

a + 3(-3) = -15

a - 9 = -15

a = -15 + 9

a = -6

Sum of the first 15 terms

S = n/2[2a + (n − 1) × d]

= 15/2 {2×-6 + (15-1)-3}

= 7.5{-12 + (14)-3}

= 7.5{ -12 - 42}

= 7.5{-54}

= -405

Sum of the first 15 terms = -405

7 0

3 years ago

Slove the simoltanous equation by using substitution or elimination method 5x-3y=1 3x+2y=1<br>

xxTIMURxx [149]

<u>By substitution method </u>

5x - 3y =1

3x + 2y = 1

_____o____o____

$5x - 3y = 1 \\ 5x = 1 + 3y \\ x = \frac{1 + 3y}{5}$

______o____o_____

$3x + 2y = 1 \\ 3( \frac{1 + 3y}{5} ) + 2y = 1 \\ \frac{3 + 9y}{5} + 2y = 1 \\ \frac{3 +9y + 10y}{5} = 1 \\ \frac{3 + 19y}{5} = 1 \\ 3 + 19y = (1)(5) \\ 3 + 19y = 5 \\ 19y = 5 - 3 \\ 19y = 2 \\ y = \frac{2}{19}$

______o_____o____

$x = \frac{1 + 3y}{5} = \frac{1 + 3 \frac{2}{19} }{5} \\ x = \frac{1 + \frac{6}{19} }{5} = \frac{ \frac{19 + 6}{19} }{5} = \frac{ \frac{25}{19} }{5} \\ x = \frac{25}{19} \div 5 \\ x = \frac{25}{19} \times \frac{1}{5} \\ x = \frac{5}{19}$

I hope I helped you^_^

7 0

3 years ago