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blondinia [14]
3 years ago
15

A hair salon charges fixed rate of $25.00 for a haircut and then an additional $15.00 for any other services. Write a function t

o model cost of services there and then determine how many services you had if you were charged for $115.00
Mathematics
1 answer:
sergij07 [2.7K]3 years ago
8 0
<span>Fixed Cost: $25 Additional Cost: $15 per service To demonstrate how much a customer would have to pay for a trip to the salon, one can use the equation Total Cost = Fixed Cost + (Additional Cost x Number of services), or Total Cost = 25 + (15 x Number of services). If I were charged for $115, I would use the equation 115 = 25+(15xServices), and solve this to find that the number of services received was six.</span>
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Holly earns $30 for babysitting 5 hours. She earns $28 when she does chores for 4 hours. Which compares the unit
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Answer:

The unit rate for doing chores is $1 per hour more than the unit rate for babysitting

Step-by-step explanation:

30 / 5 = 6

28 / 4 = 7                             the person above me was right

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3 years ago
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A. milliliters     B. 8 fluid ounces

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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

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\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

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