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kozerog [31]
3 years ago
5

the base of a right triangle is (x-5). The height of the triangle is (2x+1). The perimeter of the triangle is (7x+5). Find the a

rea of the triangle and the length of the third side.
Mathematics
1 answer:
Dovator [93]3 years ago
7 0

Answer:

Area: (x^2)-(9x/2)-2                3rd side=4x+9

Step-by-step explanation:

First, let's find the third side. If all of the sums add up to 7x+5, then we can easily subtract the two sides that we do have.

2x+1+x-5=3x-4

Now, you have to ask yourself what added to 3x-4 can be equal to 7x+5.

It's 4x+9 because we combine like terms. That's the third side. Now, we should follow the formula for getting the area of triangle using the side lengths:

(2x+1)(x-5)/2=area of triangle

2x*x=2x^2

2x*-5=-10x

x*1=x

-5*1=-5

Now the total is 2x^2-9x-4 when you continue to combine like terms. Now all we have to do is divide by 2. So, the area is 2x^2-9x-4/2. You can simplify that. Answer is up top.

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Answer:

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In other words:

1) Multiply that number by 100.

2) Add the percent sign % to it.

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Pls help me and thank ya!!
algol13

Answer:

Option 1

Step-by-step explanation:

Rational numbers do not have infinite decimals.

Option 1:

-5, 3/4 and √49 are all rational. Correct

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5 0
2 years ago
За – 6а + 2 = 8а + 20 – 5а
Luda [366]

Answer:

a = -3

Step-by-step explanation:

3a - 6a + 2 = 8a + 20 - 5a

Combine Like Terms

-3a + 2 = 3a + 20

       -2            -2

-------------------------------------

-3a = 3a + 18

-3a  -3a

-------------------------------------

-6a = 18

----   ----

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4 0
2 years ago
Perform the indicated row operations, then write the new matrix.
Studentka2010 [4]

The matrix is not properly formatted.

However, I'm able to rearrange the question as:

\left[\begin{array}{ccc}1&1&1|-1\\-2&3&5|3\\3&2&4|1\end{array}\right]

Operations:

2R_1 + R_2 ->R_2

-3R_1 +R_3 ->R_3

Please note that the above may not reflect the original question. However, you should be able to implement my steps in your question.

Answer:

\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]

Step-by-step explanation:

The first operation:

2R_1 + R_2 ->R_2

This means that the new second row (R2) is derived by:

Multiplying the first row (R1) by 2; add this to the second row

The row 1 elements are:

\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]

Multiply by 2

2 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}2&2&2|-2\end{array}\right]

Add to row 2 elements are: \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]

\left[\begin{array}{ccc}2&2&2|-2\end{array}\right] + \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]

\left[\begin{array}{ccc}0&5&7|1\end{array}\right]

The second operation:

-3R_1 +R_3 ->R_3

This means that the new third row (R3) is derived by:

Multiplying the first row (R1) by -3; add this to the third row

The row 1 elements are:

\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]

Multiply by -3

-3 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right]

Add to row 2 elements are: \left[\begin{array}{ccc}3&2&4|1\end{array}\right]

\left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right] + \left[\begin{array}{ccc}3&2&4|1\end{array}\right]

\left[\begin{array}{ccc}0&-1&1|4\end{array}\right]

Hence, the new matrix is:

\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]

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