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tamaranim1 [39]
3 years ago
14

What is the answer to (x-5)-(4x+2)

Mathematics
1 answer:
vekshin13 years ago
6 0

Let's me simply step-by-step.

Distribute the Negative sign:

x-5-(4x+2)

x-5-1(4x+2)

x-5-4x-2

(x-4x)+(-5-2)

-3x-7

the answer was -3x-7

and I hope that works you

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Answer:

37.0149 kilometers

Step-by-step explanation:

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Help! due in 2hrs please show work :)
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Which of the following are true statements.
Mariulka [41]

Answer:

Second statement is true.

The lengths 7, 40 and 41 can not be sides of a right triangle. The lengths 12, 16, and 20 can be sides of a right triangle.

Step-by-step explanation:

for first part of statement

The lengths 7, 40 and 41 can not be sides of a right triangle.

If the square of long side is equal to the sum of square of other two sides

then the given length can be sides of a right triangle.

Check the given length by Pythagoras Theorem.

c^{2} =a^{2} +b^{2}----------(1)

Let c=41 and a = 7 and b=40

Put all the value in equation 1.

41^{2} =7^{2} +40^{2}

1681=49+1600

1681=1649

Therefore, the square of long side is not equal to the sum of square of other two sides, So given lengths 7, 40 and 41 can not be sides of a right triangle.

for second part of statement.

The lengths 12, 16, and 20 can be sides of a right triangle.

Check the given length by Pythagoras Theorem.

Let c=20 and a = 12 and b=16

20^{2} =12^{2} +16^{2}

400=144+256

400=400

Therefore, the square of long side is equal to the sum of square of other two sides, So given the lengths 12, 16, and 20 can be sides of a right triangle.

Therefore, The lengths 7, 40 and 41 can not be sides of a right triangle. The lengths 12, 16, and 20 can be sides of a right triangle.

8 0
3 years ago
Integrate Sec (4x - 1) Tan (4x - 1) Dx ​
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\bf \displaystyle\int~\cfrac{~~\begin{matrix} sin(4x-1) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }{u^2}\cdot \cfrac{du}{-4~~\begin{matrix} sin(4x-1) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\implies -\cfrac{1}{4}\int\cfrac{1}{u^2}du\implies -\cfrac{1}{4}\int u^{-2}du \\\\\\ -\cfrac{1}{4}\cdot \cfrac{u^{-2+1}}{-1}\implies \cfrac{1}{4}\cdot u^{-1}\implies \cfrac{1}{4u}\implies \cfrac{1}{4cos(4x+1)}+C

5 0
4 years ago
When you were born, your uncle put $1000 into a bank account for you. According to the terms of the account, your investment
kompoz [17]

Answer:

D. Time; amount;  $2800

I hope this is right because I was looking for the answer to the same question and this is the solution I came up with.

Step-by-step explanation:

Time is the independent variable because it is the variable that is changed.

The Amount (dollars) is the dependent variable because it is effected by the independent variable.

4 0
3 years ago
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