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3 years ago

Determine the slope of the line that passes through the points (-5, -4) and (0, 10).

Mathematics

2 answers:

svetlana [45]3 years ago

6 0

so the equation for slope is y2-y1/x2-x2

so this would be -4-10/-5-0 which equals -14/-5 which is the same thing as 14/5

so the answer is 14/5 or if its mixed then 2 4/5

Hope this helps!

pashok25 [27]3 years ago

6 0

m=y2-y1/x2-x1

m=1-1/10-(-5)

m=1-1/10 + 5

m=0/15

Slope: m=0

You might be interested in

The slope of the rafter is 15 m.Half the run of the rafter measure 12m.find the height of the ridge from the base

Korvikt [17]

Answer:

$9\; \rm m$ .

Step-by-step explanation:

Assume that the run of this rafter is level. Then the height of the ridge (the line with a question mark next to it in the diagram) should be perpendicular to the line marked with $\rm 12\; m$ . The three labelled lines in this diagram will form a right triangle.

The line marked as $15\; \rm m$ will be the hypotenuse of this right triangle.
The line marked as $12\; \rm m$ will be one of the triangle's legs.
The line representing the height of the ridge (the one with the question mark) will be the other leg of this right triangle.

Hence, the height of this ridge can be found with the Pythagorean Theorem. By the Pythagorean Theorem:

$(\text{First Leg})^2 + (\text{Second Leg})^2 = (\text{Hypotenuse})^2$ .

In this particular right triangle:

$(\text{Height})^2 + (12\; \rm m)^2 = (15\; \rm m)^2$ .

$(\text{Height})^2 = (15\; \rm m)^2 - (12\; \rm m)^2$ .

Therefore, the height of this ridge would be $\sqrt{81}\; \rm m = 9\; \rm m$ . (Note the unit.)

4 0

3 years ago

Each day a commuter takes a bus to work, the transportation system has a phone app that tells her what time the bus will arrive.

Paraphin [41]

Answer:

Step-by-step explanation:

Hello!

The commuter is interested in testing if the arrival time showed in the phone app is the same, or similar to the arrival time in real life.

For this, she piked 24 random times for 6 weeks and measured the difference between the actual arrival time and the app estimated time.

The established variable has a normal distribution with a standard deviation of σ= 2 min.

From the taken sample an average time difference of X[bar]= 0.77 was obtained.

If the app is correct, the true mean should be around cero, symbolically: μ=0

a. The hypotheses are:

H₀:μ=0

H₁:μ≠0

b. This test is a one-sample test for the population mean. To be able to do it you need the study variable to be at least normal. It is informed in the test that the population is normal, so the variable "difference between actual arrival time and estimated arrival time" has a normal distribution and the population variance is known, so you can conduct the test using the standard normal distribution.

$Z_{H_0}= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } }$

$Z_{H_0}= \frac{0.77-0}{\frac{2}{\sqrt{24} } }= 1.89$

d. This hypothesis test is two-tailed and so is the p-value.

p-value: P(Z≤-1.89)+P(Z≥1.89)= P(Z≤-1.89)+(1 - P(Z≤1.89))= 0.029 + (1 - 0.971)= 0.058

e. 90% CI

$Z_{1-\alpha /2}= Z_{0.95}= 1.645$

X[bar] ± $Z_{1-\alpha /2}* (\frac{Sigma}{\sqrt{n} } )$

0.77 ± 1.645 * $(\frac{2}{\sqrt{24} } )$

[0.098;1.442]

I hope this helps!

4 0

3 years ago

Someone please explain how to get the answer and what the answer is thank you

Natalka [10]

The volume of a cylinder is

V=hpr^2 in this case h=40, r=20/2=10, and we will approximate pi as 3.14 so:

V=40(3.14)(10^2)

V=12560 ft^3

7 0

4 years ago

Bakery makes cupcakes and three different flavors chocolate, vanilla, and strawberry, with two choices of topics, butterscotch o

muminat

Answer:

24 or 12

Step-by-step explanation:

Multiplication

7 0

3 years ago

The shadow of a tower at a time is three times as long as its shadow when the angle of elevation of the Sun is 60°. Find the ang

Naddika [18.5K]

Answer:

$30^{\circ}$ .

Step-by-step explanation:

Let $\theta$ denote the unknown angle of elevation. Let $h$ denote the height of the tower.

Refer to the diagram attached. In this diagram, ${\sf A}$ denotes the top of the tower while ${\sf B}$ denote the base of the tower; ${\sf BC}$ and ${\sf BD}$ denote the shadows of the tower when the angle of elevation of the sun is $60^{\circ}$ and $\theta$ , respectively. The length of segment ${\sf AB}$ is $h$ ; $\angle {\sf ACB} = 60^{\circ}$ , $\angle {\sf ADB} = \theta$ , and ${\sf BD} = 3\, {\sf BC}$ ..

Note that in right triangle $\triangle {\sf ABC}$ , segment ${\sf AB}$ (the tower) is opposite to $\angle {\sf ACB}$ . At the same time, segment ${\sf BC}$ (shadow of the tower when the angle of elevation of the sun is $60^{\circ}$ ) is adjacent to $\angle {\sf ACB}$ .

Thus, the ratio between the length of these two segments could be described with the tangent of $m\angle {\sf ACB} = 60^{\circ}$ :

$\begin{aligned}\tan(\angle {\sf ACB}) &= \frac{\text{opposite}}{\text{adjacent}} = \frac{{\sf AB}}{{\sf BC}}\end{aligned}$ .

$\begin{aligned}\frac{{\sf AB}}{{\sf BC}} = \tan(60^{\circ}) = \sqrt{3}\end{aligned}$ .

The length of segment ${\sf AB}$ is $h$ . Rearrange this equation to find the length of segment ${\sf BC}$ :

$\begin{aligned} {\sf BC} &= \frac{{\sf AB}}{\tan(\angle ACB)} \\ &= \frac{h}{\tan(60^{\circ})}\\ &= \frac{h}{\sqrt{3}} \\ &\end{aligned}$ .

Therefore:

$\begin{aligned}{\sf BD} &= 3\, {\sf BC} \\ &= \frac{3\, h}{\sqrt{3}} \\ &= (\sqrt{3})\, h\end{aligned}$ .

Similarly, in right triangle ${\sf ABD}$ , segment ${\sf AB}$ (the tower) is opposite to $\angle {\sf ADB}$ . Segment ${\sf BD}$ (shadow of the tower, with $\theta$ as the angle of elevation of the sun) is adjacent to $\angle {\sf ADB}$ .

$\begin{aligned}\tan(\angle {\sf ADB}) &= \frac{\text{opposite}}{\text{adjacent}} = \frac{{\sf AD}}{{\sf BD}}\end{aligned}$ .

$\begin{aligned}\frac{{\sf AB}}{{\sf BD}} = \tan(\theta) \end{aligned}$ .

Since ${\sf AB} = h$ while ${\sf BD} = (\sqrt{3})\, h$ :

$\begin{aligned}\tan(\theta) &= \frac{{\sf AB}}{{\sf BD}} \\ &= \frac{h}{(\sqrt{3})\, h} \\ &= \frac{1}{\sqrt{3}}\end{aligned}$ .

Therefore:

$\begin{aligned}\theta &= \arctan\left(\frac{1}{\sqrt{3}}\right) \\ &= 30^{\circ}\end{aligned}$ .

In other words, the angle of elevation of the sun at the time of the longer shadow would be $30^{\circ}$ .

8 0

2 years ago