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3 years ago

Please show your work and explain it.

Mathematics

1 answer:

Maru [420]3 years ago

6 0

Answer:

$f(x)=\dfrac{x+2}{2(x-2)}$

Step-by-step explanation:

Remember when you divide fractions, you need to get the reciprocal of the divisor and multiply. So your first simplification would be:

$\dfrac{x^2+4x+4}{x^2-6x+8}\div\dfrac{6x+12}{3x-12}\\\\=\dfrac{x^2+4x+4}{x^2-6x+8}\times\dfrac{3x-12}{6x+12}\\\\=\dfrac{(x^2+4x+4)(3x-12)}{(x^2-6x+8)(6x+12)}$

Next we factor what we can so we can further simplify the rest of the equation:

$=\dfrac{(x^2+4x+4)(3x-12)}{(x^2-6x+8)(6x+12)}\\\\=\dfrac{(x+2)(x+2)(3x-12)}{(x^2-6x+8)(6(x+2))}\\\\$

We can now cancel out (x+2)

$=\dfrac{(x+2)(3x-12)}{(x^2-6x+8)(6)}$

Next we factor out even more:

$=\dfrac{(x+2)(3)(x-4)}{(x-2)(x-4)(6)}$

We cancel out x-4 and reduce the 3 and 6 into simpler terms:

$=\dfrac{(x+2)(1)}{(x-2)(2)}$

And we can now simplify it to:

$=\dfrac{x+2}{2(x-2)}$

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See explanation

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Theorem 2: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

1. Start point: By the 1st theorem,

$x^2=25\cdot (49-25)=25\cdot 24=5^2\cdot 2^2\cdot 6\Rightarrow x=5\cdot 2\cdot \sqrt{6}=10\sqrt{6}.$

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$x^2=40\cdot (40+5)=4\cdot 5\cdot 2\cdot 9\cdot 5\Rightarrow x=2\cdot 5\cdot 3\cdot \sqrt{2}=30\sqrt{2}.$

3. West point from the previous: By the 2nd theorem,

$x^2=(32-20)\cdot 32=4\cdot 3\cdot 16\cdot 2\Rightarrow x=2\cdot 4\cdot \sqrt{6}=8\sqrt{6}.$

4. West point from the previous: By the 1st theorem,

$9^2=x\cdot 15\Rightarrow x=\dfrac{81}{15}=\dfrac{27}{5}=5.4.$

5. West point from the previous: By the 2nd theorem,

$10^2=8\cdot (8+x)\Rightarrow 8+x=12.5,\ x=4.5.$

6. North point from the previous: By the 1st theorem,

$x^2=48\cdot 6=6\cdot 4\cdot 2\cdot 6\Rightarrow x=6\cdot 2\cdot \sqrt{2}=12\sqrt{2}.$

7. East point from the previous: By the 2nd theorem,

$x^2=22.5\cdot 30=225\cdot 3\Rightarrow x=15\sqrt{3}.$

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$x^2=12.5\cdot (12.5+13.5)=12.5\cdot 26=25\cdot 13\Rightarrow x=5\sqrt{13}.$

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$12^2=x\cdot 30\Rightarrow x=\dfrac{144}{30}=4.8.$

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$6^2=1.6\cdot (x-1.6)\Rightarrow x-1.6=22.5,\ x=24.1.$

11. East point from the previous: By the 2nd theorem,

$20^2=32\cdot (32-x)\Rightarrow 32-x=12.5,\ x=19.5.$

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The number of terms in this series is (998 - 6) / 2 + 1

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