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3 years ago

The graph of which equation has the same slope as the graph of y = 4x + 2

Mathematics

2 answers:

mel-nik [20]3 years ago

5 0

Answer:

Y = mx + b is the equation for a straight line. "B" is called the y intercept. "M" is the value of the slope of the line. "X" is the value where the line intercepts the x axis.

so the reason the awnser is D is because the M=4 in both equations, meaning that they have the same slope

Gnom [1K]3 years ago

4 0

Answer:

Step-by-step explanation:

The slope-intercept form of a line is y=mx+b where m is the slope and b is the y-intercept.

The slope of y=4x+2 is 4.

Let's look at the choices:

A) The slope of y=-2x+3 is -2.

B) The slope of y=2x-3 is 2.

C) The slope of y=-4x+2 is -4.

D) The slope of y=4x-2 is 4.

So we are looking for a line that has the same slope as the given line which is 4.

The answer is D.

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What is the value for f (x) = 4^2x -100 when x = 2

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X=2
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3 years ago

Graph the line using intercepts: 4x-3y=12

Furkat [3]

First you have to find the intercepts!

To find the y-intercept (I usually do that first), pretend x=0.
Then it becomes:
4(0)-3y=12
Which then becomes:
-3y=12.
Divide both sides, and y = -4.
Then the (x,y) coordinate would become (0, -4).

Next you would find the x-intercept, and pretend y = 0.
Which then becomes:
4x-3(0)=12
3x0 cancels out to become:
4x = 12
And x = 3 because division stuffs.
Then THIS point would then be (3,0).

So then, you would plot (0,-4) and (3,0) on a graph and connect the points! I attached a picture for reference.

hope that helped~

8 0

3 years ago

Read 2 more answers

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3 years ago

A consulting firm has received 2 Super Bowl playoff tickets from one of its clients. To be fair, the firm is randomly selecting

Soloha48 [4]

Answer:

Therefore the only statement that is not true is b.)

Step-by-step explanation:

There employees are 6 secretaries, 5 consultants and 4 partners in the firm.

a.) The probability that a secretary wins in the first draw

= $\frac{number \hspace{0.1cm} of \hspace{0.1cm} secreataries}{total \hspace{0.1cm} number \hspace{0.1cm} of \hspace{0.1cm} employees} = \frac{6}{15}$

b.) The probability that a secretary wins a ticket on second draw. It has been given that a ticket was won on the first draw by a consultant.

p(secretary wins on second draw | consultant wins on first draw)

= $\frac{p((consultant \hspace{0.1cm} wins \hspace{0.1cm} on \hspace{0.1cm} first \hspace{0.1cm}draw)\cap( secretary\hspace{0.1cm} wins\hspace{0.1cm} on second \hspace{0.1cm}draw))}{p(consultant \hspace{0.1cm} wins \hspace{0.1cm} on \hspace{0.1cm} first \hspace{0.1cm}draw)}$

= $\frac{\frac{5}{15} \times \frac{6}{14}}{\frac{5}{15} } = \frac{6}{14}$ .

The probability that a ticket was won on the first draw by a consultant a secretary wins a ticket on second draw = $\frac{6}{15}$ is not true.

The probability that a secretary wins on the second draw = $\frac{number \hspace{0.1cm} of \hspace{0.1cm} secretaries \hspace{0.1cm} remaining } { number \hspace{0.1cm} of \hspace{0.1cm} employees \hspace{0.1cm} remaining} = \frac{6 - 1}{15 - 1} = \frac{5}{14}$

c.) The probability that a consultant wins on the first draw =

$\frac{number \hspace{0.1cm} of \hspace{0.1cm} consultants \hspace{0.1cm} } { number \hspace{0.1cm} of \hspace{0.1cm} employees \hspace{0.1cm} } = \frac{5 }{15} = \frac{1}{3}$