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3 years ago

Find the slope of the line.

Mathematics

1 answer:

andrey2020 [161]3 years ago

5 0

Answer:

-1

Step-by-step explanation:

To find the slope of a line, we can do rise over run, or y/x.

r/r = 1/1 = 1

But as you can see, the line is going down which means it's decreasing, which means that it has a negative slope

slope = -1

Hope this helped a little!

Have a supercalifragilisticexpialidocious day!

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Ruben and Tariq have 245.5 downloaded minutes of

Strike441 [17]

Answer:

113.5

Step-by-step explanation:

R+T= 245.5

R=132

245.5-R=T

245.5-132=113.5

T=113.5

8 0

3 years ago

Read 2 more answers

1. Hoover Dam spans the Colorado

Alona [7]

BEST estimate of this amount is $2(10^{7})$ . Correct option A. 2 x 10^7 .

<u>Step-by-step explanation:</u>

Here we have , Hoover Dam spans the Colorado River on the Nevada-Arizona state line. For its construction, the federal government purchased 21,670,000 pounds of gates and valves. We need to find BEST estimate of this amount . Let's find out:

21,670,000 pounds

⇒ 21,670,000 pounds

⇒ $21,670,000$

⇒ $2167(10^{4})$

⇒ $2.167(10^{3})(10^{4})$

⇒ $2.167(10^{3+4})$

⇒ $2.167(10^{7})$

Now, Rounding of 2.167 we get 2 ,So

⇒ $2(10^{7})$

Hence , BEST estimate of this amount is $2(10^{7})$ . Correct option A. 2 x 10^7 .

5 0

3 years ago

Given the following right triangle, find cose, sine, tane sececsce, and cote. Do not

IgorLugansk [536]

Step-by-step explanation:

Use the Pythagorean theorem:

$leg^2+leg^2=hypotenuse^2$

We have

$leg=4,\ hypotenuse=10$

Substitute:

$4^2+leg^2=10^2$

$16+leg^2=100$ <em>subtract 16 from both sides</em>

$leg^2=84\to leg=\sqrt{84}\\\\leg=\sqrt{(4)(21)}\\\\leg=\sqrt4\cdot\sqrt{21}\\\\leg=2\sqrt{21}$

$sine=\dfrac{opposite}{hypotenuse}\\\\cosine=\dfrac{adjacent}{hypotenuse}\\\\tangent=\dfrac{opposite}{adjacent}\\\\cotangent=\dfrac{adjacent}{opposite}\\\\secant=\dfrac{hypotenuse}{adjacent}\\\\cosecant=\dfrac{hypotenuse}{opposite}$

Substitute:

$hypotenuse=10,\ opposite=4,\ adjacent=2\sqrt{21}$

$\sin\theta=\dfrac{4}{10}=\dfrac{2}{5}\\\\\cos\theta=\dfrac{2\sqrt{21}}{10}=\dfrac{\sqrt{21}}{5}\\\\\tan\theta=\dfrac{4}{2\sqrt{21}}=\dfrac{2}{\sqrt{21}}\cdot\dfrac{\sqrt{21}}{\sqrt{21}}=\dfrac{2\sqrt{21}}{21}\\\\\cot\theta=\dfrac{2\sqrt{21}}{4}=\dfrac{\sqrt{21}}{2}\\\\\sec\theta=\dfrac{10}{2\sqrt{21}}=\dfrac{5}{\sqrt{21}}\cdot\dfrac{\sqrt{21}}{\sqrt{21}}=\dfrac{5\sqrt{21}}{21}\\\\\csc\theta=\dfrac{10}{4}=\dfrac{5}{2}$