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3 years ago

What is 656,32.25 in expanded form

Mathematics

1 answer:

maks197457 [2]3 years ago

6 0

65,632.25 in expanded form is : 60,000 + 5,000 + 600 + 30 + 5 + .2 + .05

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Solve for j.<br>-13j - 20 = -8j + 20

aleksley [76]

Answer:

j = -8

Step-by-step explanation:

-13j - 20 = -8j + 20

Add 13 j to each side

-13j+13j - 20 = -8j+13j + 20

-20 = 5j+20

subtract 20 from each side

-20-20 = 5j +20-20

-40 = 5j

Divide by 5

-40/5 = =5j/5

-8 =j

8 0

3 years ago

Lorenz total payment for his loan was $13,536.

Vera_Pavlovna [14]

Answer: $376

Step-by-step explanation:

Divide 13,536 by 36

13,536/36 = 376

6 0

4 years ago

Read 2 more answers

a business want to increase its number of employees by 65% to 20,460. How many employees does the company currently have?

Crazy boy [7]

Answer:

12,400

Step-by-step explanation:

Let the current employees be x

65% increase in number = 0.65x

If the total number of employees after the increase is 20,460, then;

x + 0.65x = 20,460

1.65x = 20,460

x = 20,460/1.65

x = 12,400

Hence the current number of employees is 12,400

5 0

3 years ago

A company pays a bonus to four employees A, B, C, and D. A gets four times as much as B. B gets 50% of the amount paid to C. C a

Tomtit [17]

Answer:

A = 800, B = 200, C = 400 Andy D = 400

Step-by-step explanation:

5 0

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