All categories

4 years ago

The ratio of boys to girls at a movie is 8:7. If there are 21 girls, how many boys are at the movie?

Mathematics

1 answer:

Vera_Pavlovna [14]4 years ago

6 0

Answer:

24 boys

Step-by-step explanation:

You might be interested in

The moxy cab company charges $2 per cab ride, plus an additional $3 per mile. write an equation to represent this relationship.

S_A_V [24]

2+3m where m is miles

3 0

4 years ago

HELP ASAP PLEASEE!!

Pani-rosa [81]

Answer:

5^6 × 5^(-2) × 5^(-7)

= 5^{6+(-2)+(-7)}

= 5^{6-2-7}

= 5^(-3)

Option c 5^-3

8 0

3 years ago

Part B: The area of a rectangle is (81x2 - 4y2) square units. Determine the dimensions of the

DanielleElmas [232]

Answer:

thanksss

Step-by-step explanation:

5 0

3 years ago

Use the given transformation to evaluate the integral. double integral 9xy dA R , where R is the region in the first quadrant bo

lianna [129]

It looks like the boundaries of $R$ are the lines $y=\dfrac23x$ and $y=3x$ , as well as the hyperbolas $xy=\frac23$ and $xy=3$ . Naturally, the domain of integration is the set

$R = \left\{(x,y) ~:~ \dfrac{2x}3 \le y \le 3x \text{ and } \dfrac23 \le xy \le 3 \right\}$

By substituting $x=\frac uv$ and $y=v$ , so $xy=u$ , we have

$\dfrac23 \le xy \le 3 \implies \dfrac23 \le u \le 3$

and

$\dfrac{2x}3 \le y \le 3x \implies \dfrac{2u}{3v} \le v \le \dfrac{3u}v \implies \dfrac{2u}3 \le v^2 \le 3u \implies \sqrt{\dfrac{2u}3} \le v \le \sqrt{3u}$

so that

$R = \left\{(u,v) ~:~ \dfrac23 \le u \le 3 \text{ and } \sqrt{\dfrac{2u}3 \le v \le \sqrt{3u}\right\}$

Compute the Jacobian for this transformation and its determinant.

$J = \begin{bmatrix}x_u & x_v \\ y_u & y_v\end{bmatrix} = \begin{bmatrix}\dfrac1v & -\dfrac u{v^2} \\\\ 0 & 1 \end{bmatrix} \implies \det(J) = \dfrac1v$

Then the area element under this change of variables is

$dA = dx\,dy = \dfrac{du\,dv}v$

and the integral transforms to

$\displaystyle \iint_R 9xy \, dA = \int_{2/3}^3 \int_{\sqrt{2u/3}}^{\sqrt{3u}} \frac{dv\,du}v$

Now compute it.

$\displaystyle \iint_R 9xy \, dA = \int_{2/3}^3 \ln|v|\bigg|_{v=\sqrt{2u/3}}^{v=\sqrt{3u}} \,du \\\\ ~~~~~~~~ = \int_{2/3}^3 \ln\left(\sqrt{3u}\right) - \ln\left(\sqrt{\frac{2u}3}\right) \, du \\\\ ~~~~~~~~ = \frac12 \int_{2/3}^3 \ln(3u) - \ln\left(\frac{2u}3\right) \, du \\\\ ~~~~~~~~ = \frac12 \int_{2/3}^3 \ln\left(\frac{3u}{\frac{2u}3}\right) \, du \\\\ ~~~~~~~~ = \frac12 \ln\left(\frac92\right) \int_{2/3}^3 du \\\\ ~~~~~~~~ = \frac12 \ln\left(\frac92\right) \left(3-\frac23\right) = \boxed{\frac76 \ln\left(\frac92\right)}$

7 0

2 years ago

Im having trouble with this question, ill give 20 points :)

maks197457 [2]

Answer:

Step-by-step explanation:

the correct answer is C

3 0

3 years ago

Read 2 more answers