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3 years ago

15

Randolph is trying to find the equation of a line parallel to y = '1 over 4 x − 6 in slope-intercept form that passes through th

e point (−1, 5). Which of the following equations will he use? (6 points)

2 answers:

Gnom [1K]3 years ago

7 0

Answer:

y - 5 = $\frac{1}{4}$ (x - (-1)) The First Option

Step-by-step explanation:

if your on question 13 of (04.02 MC)

then that should be the answer.

Aleksandr [31]3 years ago

4 0

So remember that paralell lines have the same slope
y=1/4x-6
1/4=slope
so we set up a new equation in slope intercept form
y=mx+b
m=slope
b=y-intercept

y=1/4x+b

one solution si (-1,5) where x=-1 and y=5
subsitute
5=1/4(-1)+b
5=-1/4+b
add 1/4 to both sides
5+1/4=21/4=b

the answer is y=1/4x+21/4

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X^(log_(\sqrt(x))(x-3))=4

Zina [86]

Answer:

X= -4

Step-by-step explanation:

3 0

3 years ago

Find the cube roots of 125(cos 288° + i sin 288°)

kow [346]

Let r(cos O + i sin O) be a cube root of 125(cos 288 + i sin 288)
then
r^3(cos O + i sin O)^3 = 125(cos 288 + i sin 28)

so r^3 = 125 and cos 3O + i sin 3O = cos 288 + i sin 288

so r = 5 and 3O = 288 + 360p and O = 96 + 120p

so one cube root is 5 (cos 96 + i sin 96)

Im a little rusty at this stuff Its been a long time.

Im not sure of the other 2 roots

sorry cant help you any more

3 0

3 years ago

Read 2 more answers

What are the next few terms after the following sequence? 1, 1/2, 1/3, 1/4

DerKrebs [107]

Answer:

1/5, 1/6, 1/7, 1/8,1/9, 1/10,1/11,1/12

3 0

3 years ago

Solve the equation<br> (0)<br> 36<br> (6)<br> O A * = 8<br> O B x=7<br> 0 CX=512<br> 0 D<br> 3

liraira [26]

Answer

A.x=8

1/3x-2/3x=2

x-2=6

x=6x2

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4 0

3 years ago

Find all points on the circle (x^2) + (y^2) = 676 where the slope is 5/12

borishaifa [10]

Implicit defferentation

remember that dy/dx y= dy/dx

so
take derivitive of both sides
$2x+2y \space\ \frac{dy}{dx}=0$
solve for $\frac{dy}{dx}$
minus 2x both sides
$2y \space\ \frac{dy}{dx}=-2x$
divide both sides by 2y
$\frac{dy}{dx}=\frac{-2x}{2y}$
$\frac{dy}{dx}=\frac{-x}{y}$

when is the slope equal to $\frac{5}{12}$
solve for $\frac{dy}{dx}=\frac{5}{12}$
$\frac{dy}{dx}=\frac{5}{12}$
$\frac{5}{12}=\frac{-x}{y}$
$5y=-12x$
$y=\frac{-12}{5}x$
find where the circle and this line intersects

substitute $\frac{-12}{5}x$ for y
$x^2+(\frac{-12}{5}x)^2=676$
$x^2+\frac{144}{25}x^2=676$
$\frac{25}{25}x^2+\frac{144}{25}x^2=676$
$\frac{169}{25}x^2=676$
times both sides by $\frac{25}{169}$
$x^2=100$
sqrt both sides, take positive and negative roots
x=+/-10

sub back

$y=\frac{-12}{5}x$
$y=(\frac{-12}{5})(10) \space\ or \space\ (\frac{-12}{5})(-10)$
$y=\frac{-120}{5} \space\ or \space\ \frac{120}{5}$
$y=-24 \space\ or \space\ 24$

the points are (10,-24) and (-10,24)

8 0

3 years ago