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densk [106]
4 years ago
5

Solve for x x + 1/2 = -5/8 simplify your answer as much as possible.

Mathematics
1 answer:
luda_lava [24]4 years ago
5 0

Answer:

x =  -  \frac{9}{8}

Step-by-step explanation:

x +  \frac{1}{2}  =   - \frac{5}{8}

\frac{1}{2}  \times  \frac{4}{4}  =  \frac{4}{8}

x +  \frac{4}{8}  =  -  \frac{5}{8}

x +  \frac{4}{8}  -  \frac{4}{8}  =  -  \frac{5}{8}  -  \frac{4}{8}

x =  -  \frac{9}{8}

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4 years ago
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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}

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

4 0
3 years ago
What number is equivilent to 3 4/ 3 2
ki77a [65]

Answer:

9

Step-by-step explanation:

Using the property that  

a

m

a

n

=

a

m

−

n

, we have

3

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2

2

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3

4

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2

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Note that if we evaluated the numerator and denominator first, we would arrive at the same result:

3

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2

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9

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This gives

n² + 13n + 40 = (n + 8)(n + 5)

Hence, the equation represented by Ms. Wilson's model is n² + 13n + 40 = (n + 8)(n + 5)

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