Ask question

Ask question

All categories

2 years ago

6

which is the equation of a line that passes through the point (8, 2) and is perpendicular to the line y = -1/4x + 2? A y = -1/4x

B y= -1/4x + 4 C y = 4x - 30

1 answer:

Ymorist [56]2 years ago

4 0

Answer: C

Step-by-step explanation:

If a line is perpen to another line, then the other line has the opposite reciprocal of the slope.

y=-(1/4)x+2

slope of the other equation is +4

oly answer with positive slope is C

You might be interested in

What is the y intercept?

Naddika [18.5K]

Answer:

y int is 20

slope = 2

equation: y = 2x + 20

7 0

2 years ago

WRITING TO EXPLAIN, HOW COULD YOU USE 5 X 6 =30 TO FIND THE PRODUCT <br> OF 6 X6 ?

Solnce55 [7]

5x6=30
6x6=36

You would need 1 more 6
I can't explain this very well, sorry.

8 0

3 years ago

Read 2 more answers

What should the next lette be? A Z E B I Y O

myrzilka [38]

The letter C, because the pattern is this (I think):

Every two letters it states the vowels (a,e,i,o,u)

Then in between every two letters there starts a pattern of, saying the alphabet in order and saying the alphabet backwards.

In other words they merged these two lines:

A, B, C, D, E...

+

Z, Y, X, W, V...

=

A, Z, B, Y, C, X...etc.

then by adding this pattern of vowels you get:

A, Z, E, B, I, Y, O, (C), (U), (X)...

7 0

2 years ago

PLEASE HELP ME NOW URGENT

timama [110]

ANSWER

$y = 3 \pm \sqrt{21}$

EXPLANATION

The quadratic equation is:

${y}^{2} - 6y - 12 = 0$

Group variable terms:

${y}^{2} - 6y = 12$

Add the square of half, the coefficient of y to both sides.

${y}^{2} - 6y + ( - 3) ^{2} = 12 + ( - 3) ^{2}$

${y}^{2} - 6y + 9= 12 + 9$

The LHS us now a perfect square trinomial:

${(y - 3)}^{2}= 21$

Take square root:

$y - 3 = \pm \sqrt{21}$

$y = 3 \pm \sqrt{21}$

The first choice is correct.

8 0

3 years ago

Read 2 more answers

2(x+1)=3x-x+2 help fast and show work please

Stells [14]

Answer:

0=0

Step-by-step explanation:

$2(x+1)=3x-x+2\\\\\mathrm{Add\:similar\:elements:}\:3x-x=2x\\2\left(x+1\right)=2x+2\\\\\mathrm{Expand\:}2\left(x+1\right):\quad 2x+2\\2x+2=2x+2\\\\\mathrm{Subtract\:}2\mathrm{\:from\:both\:sides}\\2x+2-2=2x+2-2\\\\Simplify\\2x=2x\\\\\mathrm{Subtract\:}2x\mathrm{\:from\:both\:sides}\\2x-2x=2x-2x\\\\Simplify\\0=0\\\\\mathrm{True\:for\:all}\:x$

5 0

3 years ago