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zloy xaker [14]
3 years ago
15

What is the area of right triangle ABC where the dimensions of the two legs

Mathematics
1 answer:
Thepotemich [5.8K]3 years ago
4 0

Answer:

3x^2+17/2x-7

Step-by-step explanation:

area = 1/2 base * height

           1/2 (2x+7) (3x-2)  you must distribute first.

6x^2-4x+21x-14

3x^2 +17/2x-7

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What’s the median of these two numbers ?<br> 13 and 13
ale4655 [162]

median: The number that occurs in the middle after arranging in ascending order of magnitude

;13 13

since both appeared in the middle,you add it and divide by 2

13+13/2

26/2

13

;The median is 13

8 0
3 years ago
I need help with 27 and 28 show step and round to nearest tenth
Nikitich [7]

Answer:

#27: 56 ft squared #28: 32.85 ft squared

Step-by-step explanation:

Hope this helped! Please rate it if this helped!

5 0
3 years ago
...........................
9966 [12]
The answer to your question would be A
7 0
3 years ago
What factor is paired with 5 to give 45
Elena L [17]
9 9x5=45

5,10,15,20,25,30,35,40,45
                 9
7 0
3 years ago
Solve for x in the equation x2 - 4x-9 = 29.
erastova [34]

Answer:

x=2+\sqrt{21}\\\\x=2-\sqrt{21}

Step-by-step explanation:

One is given the following equation;

x^2-4x-9=29

The problem asks one to find the roots of the equation. The roots of a quadratic equation are the (x-coordinate) of the points where the graph of the equation intersects the x-axis. In essence, the zeros of the equation, these values can be found using the quadratic formula. In order to do this, one has to ensure that one side of the equation is solved for (0) and in standard form. This can be done with inverse operations;

x^2-4x-9=29

x^2-4x-38=0

This equation is now in standard form. The standard form of a quadratic equation complies with the following format;

ax^2+bx+c

The quadratic formula uses the coefficients of the quadratic equation to find the zeros this equation is as follows,

\frac{-b(+-)\sqrt{b^2-4ac}}{2a}

Substitute the coefficients of the given equation in and solve for the roots;

\frac{-(-4)(+-)\sqrt{(-4)^2-4(1)(-38)}}{2(1)}

Simplify,

\frac{-(-4)(+-)\sqrt{(-4)^2-4(1)(-38)}}{2(1)}\\\\=\frac{4(+-)\sqrt{16+152}}{2}\\\\=\frac{4(+-)\sqrt{168}}{2}\\\\=\frac{4(+-)2\sqrt{21}}{2}\\\\=2(+-)\sqrt{21}

Therefore, the following statement can be made;

x=2+\sqrt{21}\\\\x=2-\sqrt{21}

8 0
3 years ago
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