All categories

2 years ago

Giving out brainliest answer !!!! help me out asap please !!!!

Mathematics

1 answer:

valentinak56 [21]2 years ago

5 0

Answer:

8.07

Step-by-step explanation:

sin(48)=x/6

6sin(48)=x

You might be interested in

What problem makes 49

torisob [31]

45 + 4 makes 49 if that counts

6 0

2 years ago

On every 3rd day Ivan goes to the gym to exercise. Every 5th day Gavin goes to the gym to exercise. What is the first day Ivan a

Colt1911 [192]

You have to find the LCM (least common multiple) of both numbers.

3: 3, 6, 9, 12, 15
5: 5, 10, 15

ivan and gavin will be together at the gym on the 15th day.

8 0

2 years ago

Read 2 more answers

Tamu ran 30 minutes and burned 170 calories. PART A: Which equations can be used to find the average number of calories, c, burn

Lena [83]

Answer:

170 divided by 30 or calories divided by minutes

Step-by-step explanation:

the answer could be either but they both mean the same thing just in different words

8 0

2 years ago

If a car goes 280 miles in 4 hrs how far could it go in 6 hrs and 30 mins

Zina [86]

Answer:

280/4 = 70 miles per hour. Next, take the unit rate and multiply it by the number of hours you are trying to find. In this case, you are trying to find 6 hours and 30 minutes, but this can also be rewritten as 6.5, since 30 minutes is half an hour. So, in 6.5 hours, the car can go 455 miles

7 0

2 years ago

Sketch the domain D bounded by y = x^2, y = (1/2)x^2, and y=6x. Use a change of variables with the map x = uv, y = u^2 (for u ?

cluponka [151]

Under the given transformation, the Jacobian and its determinant are

$\begin{cases}x=uv\\y=u^2\end{cases}\implies J=\begin{bmatrix}v&u\\2u&0\end{bmatrix}\implies|\det J|=2u^2$

so that

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\iint_{D'}\frac{2u^2}{u^2}\,\mathrm du\,\mathrm dv=2\iint_{D'}\mathrm du\,\mathrm dv$

where $D'$ is the region $D$ transformed into the $u$ - $v$ plane. The remaining integral is the twice the area of $D'$ .

Now, the integral over $D$ is

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\left\{\int_0^6\int_{x^2/2}^{x^2}+\int_6^{12}\int_{x^2/2}^{6x}\right\}\frac{\mathrm dx\,\mathrm dy}y$

but through the given transformation, the boundary of $D'$ is the set of equations,

$\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}$

We require that $u>0$ , and the last equation tells us that we would also need $v>0$ . This means $1\le v\le\sqrt2$ and $0$ , so that the integral over $D'$ is

$\displaystyle2\iint_{D'}\mathrm du\,\mathrm dv=2\int_1^{\sqrt2}\int_0^{6v}\mathrm du\,\mathrm dv=\boxed6$

4 0

2 years ago