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2 years ago

It a car is traveling at 60 miles an hour how long will it take to travel 1 mile

Mathematics

1 answer:

Masja [62]2 years ago

8 0

The answer is 1 minute because an hour has 60 minutes and it jus makes sense

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Determine the image of the given point under the indicated reflection.

Aleks04 [339]

(-2,-8) is what I got after reflecting it off the y=x line.

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3 years ago

The perimeter of an equilateral triangle is 6x - 24. Which expression represents the length of one side?

PIT_PIT [208]

Answer:

side = 2x - 8

Step-by-step explanation:

An equilateral triangle has 3 equal sides.

The perimeter is 3 times one side

In this case P = 6x - 24

Side = p/3 = (6x - 24)/3

Side = 2x - 8

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3 years ago

What is the slope intercept form of the equation of the line 5x-6y=36

irina1246 [14]

It's an easy problems. what equations are you trying to find?

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3 years ago

DO U KNOW SLOPE??? PLZ HELP

IgorLugansk [536]

To calculate slope; rise/run

Triangle ABC: run-1 rise-5 slope-5

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Triangle DEF: run-2 rise 10 slope-5

10 rise/ 2 run (10/2)=5

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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