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Nesterboy [21]
3 years ago
5

Y=2x^2 Y=-3x-1 You have to solve this system of equations

Mathematics
1 answer:
skad [1K]3 years ago
8 0
y-(2*x^2)=0

y-(-3*x-1)=0
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A soccer field is a rectangle 100 m wide and 130 m long. The coach asks players to run from one corner to the other corner diago
nekit [7.7K]

Answer:

100 miles

Step-by-step explanation:

130 m and 100 m. The diagonal line would be the same length as the length of the width.  Which in this case is 100 m.

8 0
3 years ago
Read 2 more answers
What is translation?
Airida [17]

Translation is a term used in geometry to describe a function that moves an object a certain distance. The object is not altered in any other way. It is not rotated, reflected or re-sized


7 0
3 years ago
3) Malik deposited $1.050 in a savings account, and it earned $241.50 in
AnnyKZ [126]

Answer:

3%

Step-by-step explanation:

3 0
3 years ago
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
a survey showed that 82% of youth most often use the internet at home . what fraction of youth surveyed use the internet most of
Yuliya22 [10]
When talking about surveys (and most other things) we use 100% as the cap. You can not work harder than 100% efficiency and you can not have more than 100% of people agree with you. You can think of this as a pie.

A whole pie would be 100%
A half a pie 50%
Etc.

You can not have more than a whole pie (without adding another pie or variable in this case)

So when you look at this problem with that analogy in mind, all it is is simple addition and subtraction. 

If you create a survey and 82% of people vote yes, how many, out of 100%, voted yes.

To find this we take 100% (The whole pie or everybody who voted) and - the portion of the pie we ate or everyone who voted yes, 82%, which will give us the answer, everyone who voted no or the pie that is left on the platter.

So your equation would be 100 - 82 =18

18% of people interview used internet someplace other than in the home. <span />
7 0
3 years ago
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