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FromTheMoon [43]
3 years ago
5

Polygon ABCDE is reflection to produce polygon A’B’C’D E’. What is the equation for the line of reflection

Mathematics
1 answer:
lina2011 [118]3 years ago
3 0

Answer:

D Hb=HB

Step-by-step explanation:

i'm pretty sure  thats the awnser

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How do you solve this? Thank you
V125BC [204]
2)

a)

\bf a^{\frac{{ n}}{{ m}}} \implies  \sqrt[{ m}]{a^{ n}} \qquad \qquad
\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}\\\\
-------------------------------\\\\
(4x^5\cdot x^{\frac{1}{3}})+(2x^4\cdot x^{\frac{1}{3}})-(7x^3\cdot x^{\frac{1}{3}})+(3x^2\cdot x^{\frac{1}{3}})\\\\+(9x^1\cdot x^{\frac{1}{3}})-(1\cdot x^{\frac{1}{3}})
\\\\\\
4x^{5+\frac{1}{3}}+2x^{4+\frac{1}{3}}-7x^{3+\frac{1}{3}}+9x^{1+\frac{1}{3}}-x^{\frac{1}{3}}

\bf 4x^{\frac{16}{3}}+2x^{\frac{13}{3}}-7x^{\frac{10}{3}}+9x^{\frac{4}{3}}-x^{\frac{1}{3}}
\\\\\\
4\sqrt[3]{x^{16}}+2\sqrt[3]{x^{13}}-7\sqrt[3]{x^{10}}+9\sqrt[3]{x^4}-\sqrt[3]{x}

b)

\bf \cfrac{4x^5+2x^4-7x^3+3x^2+9x-1}{x^{\frac{1}{3}}}\impliedby \textit{distributing the denominator}
\\\\\\
\cfrac{4x^5}{x^{\frac{1}{3}}}+\cfrac{2x^4}{x^{\frac{1}{3}}}-\cfrac{7x^3}{x^{\frac{1}{3}}}+\cfrac{3x^2}{x^{\frac{1}{3}}}+\cfrac{9x}{x^{\frac{1}{3}}}-\cfrac{1}{x^{\frac{1}{3}}}
\\\\\\
(4x^5\cdot x^{-\frac{1}{3}})+(2x^4\cdot x^{-\frac{1}{3}})-(7x^3\cdot x^{-\frac{1}{3}})+(3x^2\cdot x^{-\frac{1}{3}})\\\\+(9x^1\cdot x^{-\frac{1}{3}})-(1\cdot x^{-\frac{1}{3}})

\bf 4x^{5-\frac{1}{3}}+2x^{4-\frac{1}{3}}-7x^{3-\frac{1}{3}}+9x^{1-\frac{1}{3}}-x^{-\frac{1}{3}}
\\\\\\
4x^{\frac{14}{3}}+2x^{\frac{11}{3}}-7x^{\frac{8}{3}}+9x^{\frac{2}{3}}-x^{-\frac{1}{3}}
\\\\\\
4\sqrt[3]{x^{14}}+2\sqrt[3]{x^{11}}-7\sqrt[3]{x^{8}}+9\sqrt[3]{x^{2}}-\frac{1}{\sqrt[3]{x}}



3)

\bf \begin{cases}
f(x)=\sqrt{x}-5x\implies &f(x)x^{\frac{1}{2}}-5x\\\\
g(x)=5x^2-2x+\sqrt[5]{x}\implies &g(x)=5x^2-2x+x^{\frac{1}{5}}
\end{cases}
\\\\\\
\textit{let's multiply the terms from f(x) by each term in g(x)}
\\\\\\
x^{\frac{1}{2}}(5x^2-2x+x^{\frac{1}{5}})\implies x^{\frac{1}{2}}5x^2-x^{\frac{1}{2}}2x+x^{\frac{1}{2}}x^{\frac{1}{5}}

\bf 5x^{\frac{1}{2}+2}-2x^{\frac{1}{2}+1}+x^{\frac{1}{2}+\frac{1}{5}}\implies \boxed{5x^{\frac{5}{2}}-2x^{\frac{3}{2}}+x^{\frac{7}{10}}}
\\\\\\
-5x(5x^2-2x+x^{\frac{1}{5}})\implies -5x5x^2-5x2x+5xx^{\frac{1}{5}}
\\\\\\
-25x^3+10x^2-5x^{1+\frac{1}{5}}\implies \boxed{-25x^3+10x^2-5x^{\frac{6}{5}}}

\bf 5\sqrt{x^5}-2\sqrt{x^3}+\sqrt[10]{x^7}-25x^3+10x^2-5\sqrt[5]{x^6}
6 0
3 years ago
If X = 6 cm and Y = 8 cm, what is the length of Z?
AleksAgata [21]
Use the trick x+ y+ z =0 and 6+8+z=0,...so z= -14
3 0
3 years ago
Solve:<br> 5x-3(10-x)&gt;it equal to 4-2x(3-x)
alexira [117]
<em><u>There is no solution!</u></em>
3 0
3 years ago
One angle within a triangle is 76 degrees. What is the value of the exterior angle?
mrs_skeptik [129]

Answer:

The value of the exterior angle is 104\°

Step-by-step explanation:

we know that

The sum of the interior angle plus the exterior angle is equal to 180\°, because are supplementary angles.

Let

x-----> the measure of the exterior angle

so

76\°+x=180\°

solve for x

x=180\°-76\°=104\°

4 0
4 years ago
Help me i dont really understand
xenn [34]

Answer:

73

Step-by-step explanation:

12(6) + 1/4 • 2^2

Square the 2 first.

12(6) + 1/4 • 4

Then multiply from left to right.

72 + 1

Lastly, add.

73

The parenthesis around the 6 mean multiplication, not the Parenthesis that are grouping symbols if you are learning PEMDAS. So your question has no grouping symbol parenthesis. Next comes Exponents, that's why we squared the 2. Multiply and Divide come next IN THE ORDER THAT THEY APPEAR FROM LEFT TO RIGHT. Then the same method with adding and subtracting. That's called the order of operations.

3 0
2 years ago
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