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3 years ago

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A line passes through the point (4,–2) and is perpendicular to the line with the equation y = 2∕3x + 4. What's the equation of t

he line? Question 1 options: A) y = –3∕2x – 4 B) y = 2∕3x + 3 C) y = 3∕2x – 8 D) y = –3∕2x + 4

1 answer:

prisoha [69]3 years ago

5 0

Answer:

Step-by-step explanation:

perp. -3/2

y + 2 = -3/2(x - 4)

y + 2 = -3/2x + 6

y = -3/2x + 4

answer is D

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What's the ordered pair ?

nekit [7.7K]

The answer will be the set of points that contains a different x value for each point.

We clearly see that for ever point in terms of CHOICE C the value of x is different.

Answer: Choice C

5 0

3 years ago

Which ordered pair makes both inequalities true? y < –x + 1 y > x On a coordinate plane, 2 straight lines are shown. The f

goldenfox [79]

Answer:

Option B.

Step-by-step explanation:

The given inequalities are

$y$

$y>x$

We need to find the ordered pair which makes both inequalities true.

Check the above inequalities for each given ordered pair.

For (-3,5),

$(5)$ (False)

For (-2,2),

$(2)$ (True)

$y>x\Rightarrow 2>-2$ (True)

So, both inequalities are true for (-2,2). Option B is correct.

For (-1,-3),

$y>x\Rightarrow -3>-1$ (False)

For (0,-1),

$y>x\Rightarrow -1>0$ (False)

Both inequalities are not true for (-3,5), (-1,-3) and (0,-1).

Therefore, the correct option is B.

3 0

2 years ago

Read 2 more answers

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

Someone please explain

Free_Kalibri [48]

Answer:

D) 6

Step-by-step explanation:

set Q has 16 values which are all even values from 4 to 36

set Z has 11 values that are multiples of 3 from 4 to 36 (6,9,12,15,etc.) however, we have to eliminate the odd ones, leaving the following six:

6, 12, 18, 24, 30, 36

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3 years ago

Read 2 more answers

The diffrence between 2 numbers is 1 and their product Is 132

il63 [147K]

Answer:

x = 12

y = 11

For any clarification, Please contact me in the comment section :)

I Hope It Helps .

Step-by-step explanation:

$Let -the-unkonown- numbers-be-x-and-y \\x-y = 1 (1)\\xy = 132 (2)\\Make-x-subject-of-formula-in-equation1 \\x = 1+y\\Substitute , 1+y for- x -in -equation ,2\\y(1+y) = 132\\y + y^{2} =132\\y^{2} +y-132=0\\Solve- the-quadratic-equation.\\y = 11\\substitute- 11 -for- y- in- equation -1- 0r- 2\\x -y = 1\\x-11 = 1\\x = 1+11\\x = 12$

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3 years ago