Please help me answer the question that’s right next to the red Transportation. Please explain how to solve as well.

Write the equation of a line perpendicular to y=3x+1and goes through the point ( 6,2) y=−13x+4 y=−13x−4 y=3x+4 y=3x−4

Korolek [52]

Answer:

$y=-\frac{1}{3}x+4$

Step-by-step explanation:

step 1

Find the slope of the perpendicular line

we know that

If two lines are perpendicular, then their slopes are opposite reciprocal

(the product of their slopes is equal to -1)

In this problem

we have

$y=3x+1$

The equation of the given line is $m=3$

so

the slope of the perpendicular line to the given line is

$m=-\frac{1}{3}$

step 2

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=-\frac{1}{3}$

$(x1,y1)=(6,2)$

substitute

$y-2=-\frac{1}{3}(x-6)$

Convert to slope intercept form

$y=mx+b$

Distribute right side

$y-2=-\frac{1}{3}x+2$

$y=-\frac{1}{3}x+2+2$

$y=-\frac{1}{3}x+4$

6 0

3 years ago

Question one* w•63=11<br>question two* 17=t•90

umka2103 [35]

#1 - 11/63
#2 - 17/90

3 0

2 years ago

Read 2 more answers

In a game of checkers there are 12 red pieces and 12 black game pieces.Julio is setting up the board to begin playing. what is t

Bond [772]

Since there are equal numbers of pieces, the chances of Julio getting a red on the first pull is 12/24 or 1/2 (reduced)

However on the second pull there are only 11 reds (Julio pulled one out) out of 23 remaining pieces so the chances of pulling the second red are 11/23

Now multiply those possibilities:
1/2 x 11/23 and you will get 11/46 which is the probability of pulling two reds in a row. 11/46 is the answer.

4 0

2 years ago

Let f (x) = 3x − 1 and ε > 0. Find a δ > 0 such that 0 < ∣x − 5∣ < δ implies ∣f (x) − 14∣ < ε. (Find the largest

s344n2d4d5 [400]

Answer:

This proves that f is continous at x=5.

Step-by-step explanation:

Taking f(x) = 3x-1 and $\varepsilon>0$ , we want to find a $\delta$ such that $|f(x)-14|$

At first, we will assume that this delta exists and we will try to figure out its value.

Suppose that $|x-5|$ . Then

$|f(x)-14| = |3x-1-14| = |3x-15|=|3(x-5)| = 3|x-5|< 3\delta$ .

Then, if $|x-5|$ , then $|f(x)-14|$ . So, in this case, if $3\delta \leq \varepsilon$ we get that $|f(x)-14|$ . The maximum value of delta is $\frac{\varepsilon}{3}$ .