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3 years ago

Y = 3x – 13 3x – 4y = 7

Mathematics

1 answer:

julsineya [31]3 years ago

8 0

What are you solving for?

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You have a bag of 42 candies you want to give to your 6 friends. the company that makes the candies guarantees that exactly 6 of

emmasim [6.3K]

The possibility is 6 out of 42

3 0

3 years ago

I need help on #10. I figured the other ones out

GalinKa [24]

First, find the ratio between the two polygons. Lets take segment MG and TV since they are congruent. MG is 4, TV is 3. The ratio of MG to TV is 4:3. We can write it as 4/3. Since other arts are congruent as well, then NK is congruent to PR, and has a ratio of 4:3. This means that (3x-2)/(x+4) is equal to 4/3. Now solve for x
(3x-2)/(x+4)=4/3
3(3x-2)=4(x+4) ---cross multiply
9x-6=4x+16
9x-4x-6=4x-4x+16 ---subtract 4x from both sides
5x-6=16
5x-6+6=16+6 ---add 6 on both sides
5x=22
5x/5=22/5 ---divide 5 on both sides
x=22/5

6 0

3 years ago

Please help, multiple choice

ycow [4]

Answer:

10

Step-by-step explanation:

5x = x20

x · x = 20 · 5

x2 = 100

x = 10

6 0

3 years ago

Read 2 more answers

Make x the subject of the formula p - qx = nx + r^2 - wx / m

Misha Larkins [42]

Answer:

x= $\frac{n-5p}{pq}$

Step-by-step explanation:

trust

6 0

3 years ago

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