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melisa1 [442]
3 years ago
13

In the figure, FL=5 and HL=8 what is FM?

Mathematics
1 answer:
liq [111]3 years ago
6 0
We know that
GL=HL/2---------> GL=8/2----> GL=4

in the right triangle FGL
FL=5
GL=4
FG=?
Applying the Pythagoras theorem
FL²=GL²+FG²-----> FG²=FL²-GL²-----> FG²=5²-4²----> FG²=9
FG=3
FG is the radius of the smaller circle
and
FM also is the radius of the smaller circle
so
FM=FG--------> FM=3

the answer is
FM is 3

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If 3/4 is the dividend and -2/5 is the divisor then what is the quotient ?
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Answer:

\dfrac{-15}{8}

Step-by-step explanation:

<h3>Fraction division:</h3>

\sf \dfrac{3}{4} \ \div \dfrac{-2}{5}

 Use KCF method.

  • Keep the first fraction.
  • Change division to multiplication
  • Flip the second fraction.

       \sf \dfrac{3}{4} \ \div \dfrac{-2}{5}=\dfrac{3}{4}*\dfrac{-5}{2}

                      \sf = \dfrac{3*(-5)}{4*2}\\\\ =\dfrac{-15}{8}

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(HURRY! I'M BEING TIMED)Write the partial fraction decomposition of the rational expression.
Aleonysh [2.5K]

Answer:

The partial fraction decomposition is \frac{- 4 x^{2} + 13 x - 12}{\left(x + 1\right)^{2} \left(x + 2\right)}=\frac{50}{x + 1}+\frac{-29}{\left(x + 1\right)^{2}}+\frac{-54}{x + 2}.

Step-by-step explanation:

Partial-fraction decomposition is the process of starting with the simplified answer and taking it back apart, of "decomposing" the final expression into its initial polynomial fractions.

To find the partial fraction decomposition of \frac{- 4 x^{2} + 13 x - 12}{\left(x + 1\right)^{2} \left(x + 2\right)}:

First, the form of the partial fraction decomposition is

                                  \frac{- 4 x^{2} + 13 x - 12}{\left(x + 1\right)^{2} \left(x + 2\right)}=\frac{A}{x + 1}+\frac{B}{\left(x + 1\right)^{2}}+\frac{C}{x + 2}

Write the right-hand side as a single fraction:

                             \frac{- 4 x^{2} + 13 x - 12}{\left(x + 1\right)^{2} \left(x + 2\right)}=\frac{\left(x + 1\right)^{2} C + \left(x + 1\right) \left(x + 2\right) A + \left(x + 2\right) B}{\left(x + 1\right)^{2} \left(x + 2\right)}

The denominators are equal, so we require the equality of the numerators:

             - 4 x^{2} + 13 x - 12=\left(x + 1\right)^{2} C + \left(x + 1\right) \left(x + 2\right) A + \left(x + 2\right) B

Expand the right-hand side:

           - 4 x^{2} + 13 x - 12=x^{2} A + x^{2} C + 3 x A + x B + 2 x C + 2 A + 2 B + C

The coefficients near the like terms should be equal, so the following system is obtained:

\begin{cases} A + C = -4\\3 A + B + 2 C = 13\\2 A + 2 B + C = -12 \end{cases}

Solving this system, we get that A=50, B=-29, C=-54.

Therefore,

                                  \frac{- 4 x^{2} + 13 x - 12}{\left(x + 1\right)^{2} \left(x + 2\right)}=\frac{50}{x + 1}+\frac{-29}{\left(x + 1\right)^{2}}+\frac{-54}{x + 2}

7 0
3 years ago
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