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3 years ago

PLEASE HELP FIRST ONE IS QUESTION THE REST ARE CHOICES Which expression is equivalent to

Mathematics

1 answer:

Readme [11.4K]3 years ago

8 0

$\sqrt[3]{256x^{10}y^7} = \sqrt[3]{4^3*4*x^{9}*x*y^6*y} = 4x^3y^2(\sqrt[3]{4xy})$

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A family has two cars. the first car has a fuel efficiency of 30 miles per gallon of gas and the second has a fuel efficiency of

jeka57 [31]

X : the gas used by first car in 1 particular week
y : the gas used by second car in 1 particular week
$x + y = 60 \\ x = 60 - y$
$30x + 35y = 1975 \\ 30 \times (60 - y) + 35y = 1975 \\ ... \\ y = 35 \\ x = 60 - y = 60 - 35 = 25$

5 0

3 years ago

F(x) = x^2 - 4, X is greater/less than 0<br><br> (a) Find the inverse function of f.<br> F^-1(x) =

Natali5045456 [20]

Step-by-step explanation:

f(x) = x² - 4

y = x²-4

x = y²-4

x + 4 = y²

$y = \sqrt{x + 4}$

7 0

2 years ago

Which graph decreases, crosses the y-axis at (0,-7), and then remains constant?

sergey [27]

Answer:

A.Graph B

Step-by-step explanation:

it decreases to a y of (0,-7) and then stays constant at -7

Hope this was helpful

7 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

Question 4(Multiple Choice Worth 5 points) (07.04 LC)What is the solution to the equation fraction 1 over 6x = 2? x = fraction 1

Elza [17]

X = fraction 1 over 12
x = fraction 1 over 3
x = 3
x = 12

The given choices are the successive steps to solve the question :fraction 1 over 6x = 2
of which the last choice is the solution.

6 0

3 years ago