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topjm [15]
3 years ago
13

Assuming the following two triangles are congruent, solve for x if QR =

Mathematics
2 answers:
makkiz [27]3 years ago
3 0

I think you meant the two <em>segments</em><em> </em>are identical, if so then:

3x + 16 = 5x - 22 \\  - 2x =  - 38 \\ x = 19

nadya68 [22]3 years ago
3 0

Answer:

x = 19

Step-by-step explanation:

3x + 16 = 5x - 22

3x - 5x = -16 - 22

-2x = -38

x = -38/-2

x = 19

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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
2. In a given population of two-earner male-female couples, male earnings have a mean of $40,000 per year and a standard deviati
Kobotan [32]

Answer: $85,000

Step-by-step explanation:

Given : In a given population of two-earner male-female couples, male earnings have a mean of $40,000 per year and a standard deviation of $12,000.

\mu_M=40,000\ \ ;\sigma_M=12,000

Female earnings have a mean of $45,000 per year and a standard deviation of $18,000.

\mu_F=45,000\ \ ;\sigma_F=18,000

If  C denote the combined earnings for a randomly selected couple.

Then, the mean of C will be :-

\mu_c=\mu_M+\mu_F\\\\=40,000+45,000=85,000

Hence, the mean of C = $85,000

8 0
3 years ago
Suppose y varies directly with x. When x is 2, y is 20. What is x when y is 50
yuradex [85]

Answer:

5

Step-by-step explanation:

x = 2 >>> (Multiply by 10) >>> y = 20

x = ? >>> (Multiply by 10) >>> y = 50

50 ÷ 10 = 5

x = 5 when y is 50

5 0
3 years ago
Fluency Practice 8x6= (8x5) + (8x__)=
Valentin [98]
The answer is 1.

8x6=48

8x5 + 8x1 = 48
4 0
3 years ago
If someone could figure out the top question that would be great, thank you so much!
Maurinko [17]
I think i only get some of it. this is what i did

x = -3 to 4y = 0.3 to 3if we use -3 and 3a < -1 < b
a= -2 to infiniteb= 0 to infinite
5 0
3 years ago
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