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3 years ago

40 percent of 15 is how much ????

Mathematics

2 answers:

kolbaska11 [484]3 years ago

8 0

Answer:

Step-by-step explanation:

40% into a decimal = .40

.40 * 15 = 6

15 - 6 = 9

timurjin [86]3 years ago

8 0

Answer:

Step-by-step explanation:

The answer is 6 because 15 x 0.4 which is 40% is 6.

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How can you find the whole when you know part and the percent.

Usimov [2.4K]

An example should make this clear.

If $50 is 40% what is the value of the whole amount?

Answer:- Whole amount = 100^%

By proportion this = (100/ 40) * 50

= 2.5 * 50

= $125 (answer)

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3 years ago

Set up an equation and solve for this problem.

viktelen [127]

Your equation might look like this,
Since the ticket is $8, and her budget is 50, your gonna have to divide.
Then the remaining answer would be, 6.25.
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4 years ago

What is 48 to the square root of 9.5

Leokris [45]

48 divided by sqrt of 9.5 =

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3 years ago

A baseball has a 48 cm diameter. What is the volume of the contents of the ball?

Andrew [12]

Answer:

201cm

Step-by-step explanation:

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3 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

3 years ago