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2 years ago

PLEASE HELP ME !!!!!!!!!!!!

Mathematics

1 answer:

Lera25 [3.4K]2 years ago

5 0

Answer:

you profit is 22 you have to divide $209 by $9.50

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If the pattern is 3ª¯², what is the value if a = 5?

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27 because 2-5=3 and 3^3 is 27

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3 years ago

How do you solve 5+4(x+9)=-3

dlinn [17]

Final answer: x=-11
1. distribute 4(x+9)
5+4x+36=-3
2. combine like terms
4x + 41 = -3
3. subtract 41 from each side
4x = -44
4. divide 4 from each side
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2 years ago

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What is 6/51 written as a lowest term

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2/17 is the lowest term for 6/51

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2 years ago

Simplify the expression to one number using order of operations:

geniusboy [140]

<h2><u>COMBINED OPERATIONS</u></h2><h3>Exercise</h3>

First, let's solve the subtraction (in parentheses):

$\mathsf{\dfrac{-8\left(2-6\right)}{2}}$

$\mathsf{\dfrac{-8(-4)}{2}}$

Now, let's solve the multiplication. Remember: in multiplication, two negative numbers will give a positive number:

$\mathsf{\dfrac{-8(-4)}{2}}$

$\mathsf{\dfrac{32}{2}}$

Finally, let's solve the division:

$\mathsf{\dfrac{32}{2} = 16}$

<h3><u>Answer</u>: 16</h3>

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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

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

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2 years ago