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hram777 [196]
2 years ago
15

Solve this PLZ! 4 - 7 (v - 9) = -7 -6(v - 10)

Mathematics
1 answer:
Arisa [49]2 years ago
7 0
I maybe wrong, but I think it is 14

Simplify both sides of the equation
(Do distributive property)
Add 6v to both sides
Subtract 67 from both sides
Divide both sides by -1
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68 I think that is the answer
7 0
3 years ago
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Convert the integral below to polar coordinates and evaluate the integral.
Lelechka [254]

Answer:

\int\limits_{0}^{4/\sqrt{2}}\int\limits_{y}^{\sqrt{16-y^2}}  xy \, dxdy =  \int\limits_{0}^{\pi/4} \, \int\limits_{0}^{4}    r^3 cos(\theta)\sin(\theta) \,\,  drd\theta = 16

Step-by-step explanation:

We are trying to evaluate this integral.

\int\limits_{0}^{4/\sqrt{2}}\,\,\int\limits_{y}^{\sqrt{16-y^2}}  xy \,\,dxdy

The first thing that we have to do is understand this region in the plane.

\{ (x,y) \in \mathbb{R} :  0\leq y \leq \frac{4}{\sqrt{2}} \,\, , y \leq x \leq \, \sqrt{16-y^2}   \}

If you graph it looks something like the photo I join.

Now we need to describe that same region in polar coordinates.

That same region in polar coordinates would be

\{ (r,\theta) : \,\, 0 \leq \theta \leq \frac{\pi}{4}  \,\,\, 0\leq r \leq 4  \}

Now remember that when we do the polar transformation we use the following formula

\int\limits_{a}^{b} \, \int\limits_{c}^{d}    f(x,y) \,dxdy =  \int\limits_{\theta_1}^{\theta_2} \, \int\limits_{r_1}^{r_2}    r* f(rcos(\theta),rsin(\theta))  \,drd\theta

Then our integral would be

\int\limits_{0}^{4/\sqrt{2}}\int\limits_{y}^{\sqrt{16-y^2}}  xy \, dxdy =  \int\limits_{0}^{\pi/4} \, \int\limits_{0}^{4}    r^3 cos(\theta)\sin(\theta) \,\,  drd\theta = 16

6 0
3 years ago
Leila's soccer team has 8 players on it. there are 40 players in the league. how many teams are playing in the league?
bazaltina [42]
There are 5 teams playing in the league
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3 years ago
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(multiple choice) plz help me!!
harkovskaia [24]

8 \sqrt{5} is \: the \: length

Step-by-step explanation:

( {16})^{2}  = ( {x})^{2}  - ( {8})^{2}  \\ 256 =  {x}^{2}  - 64 \\ 256 + 64 =  {x}^{2}  \\    \sqrt{320}  =  \sqrt{ {x}^{2} }  \\ 8 \sqrt{5}  = x

7 0
3 years ago
Pleaaaseee helpppp!​
xxMikexx [17]

Y=x2

Step-by-step explanation:

7 0
2 years ago
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