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Igoryamba
3 years ago
8

Guys!! anyone can help me solve this geometry problem?What is the area of this triangle?

Mathematics
1 answer:
NeX [460]3 years ago
3 0

Answer:

10657.5

Step-by-step explanation:

<h2>Long way that is unnecessarily long</h2>

We can start by finding the area of the larger triangle. Using the Pythagorean theorem, we can say that 251²-105²=the bottom side², and  251²-105²=51976, so the bottom side of the larger triangle is √51976 , or approximately 228. Then, the area of the larger triangle is √51976 * 105/2 = 11969 (approximately). Then, the area of the smallest triangle (the largest triangle - the one that we're trying to find the area of) is 105*(√51976-203)/2 = approximately 1312. Then, subtracting that from the total area, we get (√51976 * 105 -  105*(√51976-203))/2 = 105*203/2 = 10657.5

<h2>Short way</h2>

ALTERNATIVELY, upon further review, we can just see that the height is 105 and the base is 203, so we multiply those two and divide by 2, as is the formula for the area of a triangle, to get 10657.5

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Answer:

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Step-by-step explanation:

From the picture attached,

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Perimeter of the garden = 8 + 8 + 6

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Therefore, y = 69°

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x° + y° + 69° = 180°

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1. Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants.
german

Answer:

Step-by-step explanation:

1.

To write the form of the partial fraction decomposition of the rational expression:

We have:

\mathbf{\dfrac{8x-4}{x(x^2+1)^2}= \dfrac{A}{x}+\dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{(x^2+1)^2}}

2.

Using partial fraction decomposition to find the definite integral of:

\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}dx

By using the long division method; we have:

x^2-8x-20 | \dfrac{2x}{2x^3-16x^2-39x+20 }

                  - 2x^3 -16x^2-40x

                 <u>                                         </u>

                                            x+ 20

So;

\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x+\dfrac{x+20}{x^2-8x-20}

By using partial fraction decomposition:

\dfrac{x+20}{(x-10)(x+2)}= \dfrac{A}{x-10}+\dfrac{B}{x+2}

                         = \dfrac{A(x+2)+B(x-10)}{(x-10)(x+2)}

x + 20 = A(x + 2) + B(x - 10)

x + 20 = (A + B)x + (2A - 10B)

Now;  we have to relate like terms on both sides; we have:

A + B = 1   ;   2A - 10 B = 20

By solvong the expressions above; we have:

A = \dfrac{5}{2}     B =  \dfrac{3}{2}

Now;

\dfrac{x+20}{(x-10)(x+2)} = \dfrac{5}{2(x-10)} + \dfrac{3}{2(x+2)}

Thus;

\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)}

Now; the integral is:

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\mathbf{\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx =  x^2 + \dfrac{5}{2}In | x-10|\dfrac{3}{2} In |x+2|+C}

3. Due to the fact that the maximum words this text box can contain are 5000 words, we decided to write the solution for question 3 and upload it in an image format.

Please check to the attached image below for the solution to question number 3.

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