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3 years ago

What is the distance in units from (2, -12) to (-11, -12) 20 POINTSSS HURRY PLEASEEEE

Mathematics

2 answers:

Zinaida [17]3 years ago

5 0

Answer:

Distance = $\sqrt{(-11-2)^2 + (-12+12)^2}$

Distance = $\sqrt{(-13)^2}$

Distance = 13 units

sweet [91]3 years ago

3 0

Answer:

d= (-11-2)^2 + (-12--12)^2

d= (-13)^2 + (0)^2

d=169

square root the 169

answer is 13

that's the distance from (2, -12) (-11, -12)

hope that helps

Step-by-step explanation:

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On Tuesday, there were 400 customers at a movie theater, and each customer paid $8 for a ticket. On Wednesday, the owner of the

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In a volcano, erupting lava flows continuously through a tube system about 17 kilometers to the sea. Assume a lava flow speed of

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3 years ago

Read 2 more answers

A golfer hits a ball from a starting elevation of 2 feet with a velocity of 70 feet per second down to a green with an elevation

ikadub [295]

Answer:

4.50 s

Step-by-step explanation:

The motion of the ball is represented by the following equation:

$h(t) = 2+70t-16t^2$ (1)

where

h(t) is the elevation at time t

2 ft is the initial elevation at t = 0

+70 ft/s is the initial vertical velocity

$-32 ft/s^2$ is the acceleration due to gravity

The green is located at an elevation of -7 feet, so the ball lands on the green when

h(t) = -7

Substituting into (1)

$-7=2+70t-16t^2$

And re-arranging we get

$16t^2-70t-9=0$

This is a second-order equation in the form

$at^2+bt+c=0$

which has solutions

$t_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ (2)

Here we have:

a = 16

b = -70

c = -9

Substituting into (2) we find the solutions:

$t_{1,2}=\frac{70\pm \sqrt{(-70)^2-4(16)(-9)}}{2(16)}$

The two solutions are:

t = 4.50 s

t = -0.125 s

The second solution is negative, and since negative time has no physical meaning, the only correct solution is

t = 4.50 s

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3 years ago

A sports news station wanted to know whether people who live in the North or the South are bigger sports fans. For its study, 17

Free_Kalibri [48]

Using the z-distribution, it is found that the 95% confidence interval for the difference is (-1.3, -0.7).

<h3>What are the mean and the standard error for each sample?</h3>

Considering the data given:

$\mu_S = 3.8, n = 175, s_S = \frac{1.7}{\sqrt{175}} = 0.1285$

$\mu_N = 4.8, n = 152, s_N = \frac{1.2}{\sqrt{152}} = 0.0973$

<h3>What is the mean and the standard error for the distribution of differences?</h3>

The mean is the subtraction of the means, hence:

$\overline{x} = \mu_S - \mu_N = 3.8 - 4.8 = -1$

The standard error is the square root of the sum of the variances of each sample, hence:

$s = \sqrt{s_S^2 + s_N^2} = \sqrt{0.1285^2 + 0.0973^2} = 0.1612$

<h3>What is the confidence interval?</h3>

It is given by:

$\overline{x} \pm zs$

We have a 95% confidence interval, hence the critical value is of z = 1.96.

Then, the bounds of the interval are given as follows:

$\overline{x} - zs = -1 - 1.96(0.1612) = -1.3$

$\overline{x} + zs = -1 + 1.96(0.1612) = -0.7$

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

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2 years ago