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3 years ago

If uq=6, find us and qs

Mathematics

1 answer:

Anon25 [30]3 years ago

4 0

Can you provide more information?

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What is the slope of the line x=-3

mote1985 [20]

x = 3 is a vertical line that goes straight up and down at 3 on the x axis, vertical lines have a undefined slope. Don't confuse that with no slope which is a flat like like y=7.

so the slope is undefined

4 0

3 years ago

Read 2 more answers

Find the measure of angle C of a triangle ABC, if: m∠A = ∝, m∠B = 2∝

Effectus [21]

Answer:

The measure of ∠ C is 180°-3∝

Step-by-step explanation:

We are given a ΔABC

∠A = ∝

∠B = 2∝

Angle sum property of triangle : The sum of all angles of triangle is 180° is called angle sum property of triangle

So, ∠A+∠B+∠C=180°

∝+2∝+∠C=180°

3∝+∠C=180°

∠C=180°-3∝

Hence the measure of ∠ C is 180°-3∝

5 0

2 years ago

The zeros of function d are -3 and 8. Which expression could be a factor of function d?

marysya [2.9K]

X+3 or x-8
These answers will both work!

8 0

3 years ago

A line contains the points (-5, 3) and (0, -7). What is the equation of this line in slope-intercept form?

pychu [463]

Answer:

D. y = –2x – 7

Step-by-step explanation:

First, find the rate of change [slope]:

-y₁ + y₂\-x₁ + x₂ = m

$- \frac{3 - 7}{5 + 0} = -\frac{10}{5} = -2$

In this case, the y-intercept of [0, -7] is ALREADY given to you, so we would not have to go further in this case.

I am joyous to assist you anytime.

8 0

2 years ago

Read 2 more answers

Write an equation that represents the line. Use exact numbers. (-2,-1) (4,6)

vazorg [7]

Check the picture below.

to get the equation of any straight line, we simply need two points off of it, so let's use those in the picture

$(\stackrel{x_1}{-2}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{6})~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{6}-\stackrel{y1}{1}}}{\underset{run} {\underset{x_2}{4}-\underset{x_1}{(-2)}}} \implies \cfrac{6 -1}{4 +2} \implies \cfrac{ 5 }{ 6 }$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{1}=\stackrel{m}{\cfrac{5}{6}}(x-\stackrel{x_1}{(-2)})\implies y-1=\cfrac{5}{6}(x+2) \\\\\\ y-1=\cfrac{5}{6}x+\cfrac{5}{3}\implies y=\cfrac{5}{6}x+\cfrac{5}{3}+1\implies y=\cfrac{5}{6}x+\cfrac{8}{3}$

6 0

1 year ago