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3 years ago

Terry Wade's variable costs for his car

Mathematics

1 answer:

Pie3 years ago

8 0

Answer:

id guess if i were u but thats just me

Step-by-step explanation:

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The coffe shop made 18 lattes on Tuesday. That was three times as many as they made on Monday. The barista uses 2/3 cup of milk

Oduvanchick [21]

The barista used 16 cups of milk. I know, I know, I'm super cool, just rate me 4 stars or better, its really the only way you can repay the massive debt you owe me.

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2 years ago

EASY ALGEBRA QUESTION<br><br> Please help. This is worth 20 points and I will give brainliest!

Vadim26 [7]

Answer:

Step-by-step explanation:

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3 years ago

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Write the equation for the inverse function of y=0.4x-1.2

loris [4]

I think it is 0.4/0.4 but I'm not sure

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3 years ago

What is the area of the sector a.) 10π cm^2b.) 16π cm^2c.) 40π cm^2d.) 64π cm^2

Natalija [7]

Answer:

C. 40π cm²

Explanation:

• The central angle of the sector = 225 degrees

• The radius of the sector = 8cm

For a sector of radius r and central angle θ, we calculate the area using the formula below:

$\text{Area}=\frac{\theta}{360\degree}\times\pi r^2$

Substituting the given values, we have:

$\begin{gathered} \text{Area}=\frac{225}{360}\times\pi\times8^2 \\ =\frac{5}{8}\times64\times\pi \\ =5\times8\times\pi \\ =40\pi cm^2 \end{gathered}$

The area of the sector is 40π cm².

3 0

1 year 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

2 years ago