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3 years ago

Three times a number is greater than -15 and less than 12.

Mathematics

1 answer:

Nesterboy [21]3 years ago

6 0

Anything under 3x4 and 3x-5 is greater then -15 and less than 12.

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Plz help me plz plz thank you

DENIUS [597]

Answer:

240

Step-by-step explanation:

bruh all you have to do is (11+5)*15

6 0

3 years ago

Teesha is in French club. There are 10 freshman, 12 sophomores, 15 juniors, and 30 seniors in the club. The advisor is going to

FrozenT [24]

We know that

In French club there are
10 freshman
12 sophomores
15 juniors
30 seniors

total of the members------> (10+12+15+30)=67
total freshman-----> 10

so
the probability that a freshman will be chosen=10/67
and
the probability that a freshman will not be chosen=(67-10)/67
the probability that a freshman will not be chosen=57/67---> 0.8507
0.8507= 85.07%

the answer is
the probability that a freshman will not be chosen is 85.07%

6 0

3 years ago

The cross section of a cylinder taken parallel to the base produces which two-dimensional shape?

zysi [14]

If i had a cylinder If I cut it in right before the top, what do you think the cross section would look like? answer a circle

3 0

2 years ago

Read 2 more answers

What is the volume of the prism?

Vika [28.1K]

For the volume, the formula is length*width*height
so
(2)(1.5)(1.5)=4.5
In other words, your answer would be 4.5

8 0

2 years ago

Read 2 more answers

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