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.

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