All categories

3 years ago

Find the distance between (-4,-3) and (-2,0). Round your answer to the nearest tenth.

Mathematics

2 answers:

Kazeer [188]3 years ago

8 0

Answer:

just count the spaces between (-4, -3) (-2,0)

Step-by-step explanation:

use rise and run or sum else

dlinn [17]3 years ago

4 0

Answer:

$\sqrt{13}$

Step-by-step explanation:

$\mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$\mathrm{The\:distance\:between\:}\left(-4,\:-3\right)\mathrm{\:and\:}\left(-2,\:0\right)\mathrm{\:is\:}$

$=\sqrt{\left(-2-\left(-4\right)\right)^2+\left(0-\left(-3\right)\right)^2}$

$Simplify$

You might be interested in

Find the slope of the line passing through the points (-8, 3) and (7,3)

Aliun [14]

Answer:

0 I think??

Step-by-step explanation:

$M=\frac{x2-x1}{y2-y1}$

$M=\frac{7-(-8)}{3-3}$

$M=\frac{15}{0}$

$M=0$

5 0

3 years ago

PLEASE HELP!! SOON AS POSSIBLE.

Klio2033 [76]

2/15

90 ÷ 6 = 15

Theoretical probability would be 15 due to the equation above

7 0

3 years ago

Which set of numbers could be the length of the sides

Svetllana [295]

Answer: Choice C) {6, 9, 12}

=======================================================

Explanation:

Something like choice A can't form a triangle because the first two sides add to 6+9 = 15, which is <u>not</u> longer than the third side 15. We need to have x+y > z to be true. The same goes for choice B as well because 3+3 = 6 is too small compared to the third side 7. Choice D is similar to choice A.

In short, we can rule out choices A, B, and D.

The only thing left is choice C. Picking any two sides leads to having that sum be larger than the third side

6+9 = 15 which is larger than 12
9+15 = 24 which is larger than 6
6+15 = 21 which is larger than 9

So the conditions of the triangle inequality theorem work out here, and we have a triangle.

3 0

3 years ago

Please help!! Select the correct answer??

VladimirAG [237]

The quotient is : 5x-12+(25)/(x+3)

The remainder is : 25

5 0

4 years ago

What is the solution set for 2x+5y>-1 and 4x-3<-3?

bekas [8.4K]

PROBLEM ONE

•

Solving for x in 2x + 5y > -1.

•

Step 1 ) Subtract 5y from both sides.

2x + 5y > -1

2x + 5y - 5y > -1 - 5y

2x > -1 - 5y

Step 2 ) Divide both sides by 2.

2x > -1 - 5y

$\displaystyle\frac{2x}{2} > \displaystyle\frac{-1 - 5y}{2}$

$\displaystyle\ x > \frac{-1 - 5y}{2}$

So, the solution for x in 2x + 5y > -1 is...

$\displaystyle\ x > \frac{-1 - 5y}{2}$

•

Solving for y in 2x + 5y > -1.

•

Step 1 ) Subtract 2x from both sides.

2x + 5y > -1

2x - 2x + 5y > -1 - 2x

5y > -1 - 1x

Step 2 ) Divide both sides by 5.

5y > -1 - 1x

$\displaystyle\frac{5x}{5} > \frac{-1 -1x}{5}$

$\displaystyle\ x > \frac{-1 -1x}{5}$

So, the solution for y in 2x + 5y > -1 is...

$\displaystyle\ x > \frac{-1 -1x}{5}$

•

PROBLEM TWO

•

Solving for x in 4x - 3 < -3.

•

Step 1 ) Subtract 3 from both sides.

4x - 3 < -3

4x -3 - 3 < -3 - 3

4x < 0

Step 2 ) Divide both sides by x.

4x < 0

$\displaystyle\frac{4x}{4}$

x < 0

So, the solution for x in 4x - 3 < -3 is...

x < 0

•

- <em>Marlon Nunez</em>

7 0

3 years ago