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vaieri [72.5K]
3 years ago
7

If the slope of the line is 3, indicate whether the line through the point rises, falls, horizontal, or vertical

Mathematics
1 answer:
RSB [31]3 years ago
5 0

Answer:

the line rises from left to right

Step-by-step explanation:

the slope is a positive number, therefore it will be rising, not dropping/falling, or going down.

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Please Explain How To Solve This
atroni [7]

Answer:

Step-by-step explanation:

domain is always the x-values, the first number of an ordered pair. and yes, the x and y values increase at a constant rate.

problem #1:

x   /   y

-2     -1

-1      -3

0      -5

1       -7

problem #4:

(5, 1) (6, 2) (7, -3) (8, 4) (9, 5)

hope this all helped ;)

mark me brainliest :D

4 0
3 years ago
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Cubed root x cubed root x2​
Yuki888 [10]

Answer:

Final answer is \sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x.

Step-by-step explanation:

Given problem is \sqrt[3]{x}\cdot\sqrt[3]{x^2}.

Now we need to simplify this problem.

\sqrt[3]{x}\cdot\sqrt[3]{x^2}

\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}

Apply formula

\sqrt[n]{x^p}\cdot\sqrt[n]{x^q}=\sqrt[n]{x^{p+q}}

so we get:

\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{1+2}}

\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{3}}

\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x

Hence final answer is \sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x.

7 0
3 years ago
A candidate for city council wants to determine how much support she has among the voters in her district by conducting a survey
SpyIntel [72]
I think c .. I already took a course .. and the test had that question.. I cant remember the answer though
5 0
3 years ago
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b. Two events are dependent if the occurrence of one event changes to occurrence of the second event. True or False
Naddika [18.5K]

Answer:

true

Step-by-step explanation:

6 0
3 years ago
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A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by th
olchik [2.2K]

Answer:

14.52 seconds.

Step-by-step explanation:

We have been given that the height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation y=-16x^2+224x+121. We are asked to find the time, when the rocket will hit the ground.

We know that the rocket will hit the ground, when height will be 0. So to find the time when rocket will hit the ground, we will substitute y=0 in our given equation as:

0=-16x^2+224x+121

Let us solve for x using quadratic formula.

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

x=\frac{-224\pm\sqrt{224^2-4(-16)(121)}}{2(-16)}

x=\frac{-224\pm\sqrt{50176+7744}}{-32}

x=\frac{-224\pm\sqrt{57920}}{-32}

x=\frac{-224\pm240.66574}{-32}

x=\frac{-224+240.66574}{-32}, x=\frac{-224-240.66574}{-32}

x=\frac{16.66574}{-32}, x=\frac{-464.66574}{-32}

x=-0.520804375, x=14.520804375

Upon rounding to nearest 100th of second, we will get:

x\approx -0.52, x\approx 14.52

Since time cannot be negative, therefore, the rocket will hit the ground after 14.52 seconds.

6 0
3 years ago
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