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Mrrafil [7]
3 years ago
9

Sovle for X. Please.

Mathematics
1 answer:
Sati [7]3 years ago
5 0
Answer:
RP: x equals 0
JH: x equals -11
EG: x equals 9
You might be interested in
Use the given transformation x=4u, y=3v to evaluate the integral. ∬r4x2 da, where r is the region bounded by the ellipse x216 y2
exis [7]

The Jacobian for this transformation is

J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 3 \end{bmatrix}

with determinant |J| = 12, hence the area element becomes

dA = dx\,dy = 12 \, du\,dv

Then the integral becomes

\displaystyle \iint_{R'} 4x^2 \, dA = 768 \iint_R u^2 \, du \, dv

where R' is the unit circle,

\dfrac{x^2}{16} + \dfrac{y^2}9 = \dfrac{(4u^2)}{16} + \dfrac{(3v)^2}9 = u^2 + v^2 = 1

so that

\displaystyle 768 \iint_R u^2 \, du \, dv = 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2 \, du \, dv

Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.

\begin{cases} u = r\cos(\theta) \\ v = r\sin(\theta) \\ u^2+v^2 = r^2\\ du\,dv = r\,dr\,d\theta\end{cases}

Then

\displaystyle 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2\,du\,dv = 768 \int_0^{2\pi} \int_0^1 (r\cos(\theta))^2 r\,dr\,d\theta \\\\ ~~~~~~~~~~~~ = 768 \left(\int_0^{2\pi} \cos^2(\theta)\,d\theta\right) \left(\int_0^1 r^3\,dr\right) = \boxed{192\pi}

3 0
2 years ago
A French class has a total of 31 students. The number of males is 7 more than the number of females. How many males and how many
Inessa05 [86]

Answer:

12

Step-by-step explanation:

5 0
2 years ago
Is -27 greater then |-14|
skelet666 [1.2K]

Answer:

No

Step-by-step explanation:

the absolute value function always returns a positive value. The value inside the bars can be positive or negative but the result is always positive

| - 14 | = 14 > - 27


3 0
3 years ago
Read 2 more answers
Solve |p + 2| = 10Solve |p + 2| = 10<br><br> {-12}<br> {-8, 8}<br> {-12, 8}
valentinak56 [21]

By definition, we have

|p+2| = \begin{cases} p+2 &\text{ if } p+2 \geq 0 \\-p-2 &\text{ if } p+2 < 0 \end{cases}

So, we have to solve two different equations, depending of the possible range for the variable. We have to remember about these ranges when we decide to accept or discard the solutions:

Suppose that p+2\geq 0 \iff p \geq -2

In this case, the absolute value doesn't do anything: the equation is

p+2 = 10 \iff p = 10-2 = 8

We are supposing p \geq -2, so we can accept this solution.

Now, suppose that p+2 < 0 \iff p < -2. Now the sign of the expression is flipped by the absolute value, and the equation becomes

-p-2 = 10 \iff -p = 12 \iff p = -12

Again, the solution is coherent with the assumption, so we can accept this value as well.

3 0
3 years ago
Wath the factorization of 8
kap26 [50]
Factorization of 8
= (2*2*2)
5 0
2 years ago
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