I need help with finding slopes from two points thank you

Mathematics

1 answer:

otez555 [7]3 years ago

7 0

Answer:

i do believe the answer is 6/7

You might be interested in

Pamela bought an electric drill at 85% of the regular price. She paid $32.89 for the drill. What was the regular price? (Round t

Stolb23 [73]

<span>38.69</span>

This is because
32.89/0.85 = 38.69

5 0

3 years ago

What is d+15-9/3+5 PLZZZZZ

zloy xaker [14]

D + 15 - 9 / 3 + 5 the answer is : d + 17

3 0

2 years ago

Mathematics algebra essay

Komok [63]

Answer:

The answer is D $3.35

Step-by-step explanation:

$2.35 + .25x = 1.75 + .40x \\ 2.35 = 1.75 + .40x - .25x \\ 2.35 = 1.75 + .15x \\ 2.35 - 1.75 = .15x \\ .60 = .15x \\ \frac{.60}{.15} = x \\ x = 4$

$2.35 + .25(4) = 2.35 + 1 = 3.35$

8 0

3 years ago

Type the correct answer for each.

xxTIMURxx [149]

If point $\left(x,\ \frac{\sqrt{7}}{3}\right)$ is on the unit ircle, then:

$x^2+\left( \frac{ \sqrt{7} }{3} \right)^2=1 \\ \\ \Rightarrow x^2+ \frac{7}{9} =1 \\ \\ \Rightarrow x^2=1- \frac{7}{9} = \frac{2}{9} \\ \\ \Rightarrow x= \sqrt{ \frac{2}{9} } = \frac{ \sqrt{2} }{3}$

Since, the point is in the second quadrant, x is negative.

Thus, $(x,\ y)=\left(x,\ \frac{\sqrt{7}}{3}\right)=\left(-\frac{ \sqrt{2} }{3},\ \frac{\sqrt{7}}{3}\right)$

Part A:

$3 \sqrt{7} =6\left( \frac{ \sqrt{7} }{2} \right) \\ \\ =6\left( \frac{ \sqrt{7} }{3} \cdot \frac{3}{ \sqrt{2} } \right) \\ \\ =6\left( \frac{ \frac{ \sqrt{7} }{3} }{ \frac{ \sqrt{2} }{3} } \right)=-6\left( \frac{ \frac{ \sqrt{7} }{3} }{ -\frac{ \sqrt{2} }{3} } \right) \\ \\ =-6\left( \frac{y}{x} \right)=-6\tan\theta$

Therefore, $-6\tan\theta=3\sqrt{7}$ .

Part B:

$- \frac{ \sqrt{7} }{2} =- \frac{ \sqrt{7} }{3} \cdot \frac{3}{ \sqrt{2} } \\ \\ =- \frac{ \frac{ \sqrt{7} }{3} }{ \frac{ \sqrt{2} }{3} } = \frac{ \frac{ \sqrt{7} }{3} }{ -\frac{ \sqrt{2} }{3} } = \frac{y}{x} \\ \\ =\tan\theta$

Therefore, $\tan\theta=- \frac{ \sqrt{7} }{2}$

8 0

3 years ago

Consider the equation below.

Marta_Voda [28]

Be because I did this it’s just extremely hard to explain how to trust OK open helps

7 0

2 years ago