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Rina8888 [55]
4 years ago
9

NEED HELP ASAP!!!!!!!!!!!!!!!! If the polygon is translated 4 units down and 5 units right, what will the coordinates of the new

image be? Use prime notation in expressing the new coordinates.

Mathematics
1 answer:
Elza [17]4 years ago
7 0

Answer:

A’(-1,1), B’(-1,-2), C’(3,-2), D’(3,2)

Step-by-step explanation:

The problem is just (x+5,y-4)

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2+2+9+0+8+1+2+3+4+5+6+7+8+9 is 66
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Simplify the number into simplest radical form. Use the factor tree to help determine the factors. StartRoot 96 Endroot StartRoo
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Answer:

[tex]4608\sqrt{3}[/tex]

Step-by-step explanation:

1. \sqrt{96} *\sqrt{6}* 2*\sqrt{6}*4 *\sqrt{6} *4*\sqrt{3}

2. \sqrt{2^{5} *3} \sqrt{6} *2\sqrt{6}*4\sqrt{6}*4\sqrt{3}   <em>factoring 96</em>

<em>since \sqrt{2^{5}*3 } = \sqrt{2^{5} } \sqrt{3}</em>

3. \sqrt{2^{5} } \sqrt{3}\sqrt{6} *2\sqrt{6}*4\sqrt{6}*4\sqrt{3}

<em>using exponent rule - (a^{b}) ^{c} = a^{bc}</em>

<em> \sqrt{2^{5} } = 2^{5/2}</em>

4. 2^{5/2}\sqrt{3}\sqrt{2*3} *2\sqrt{6}*4\sqrt{6}*4\sqrt{3}

<em>doing some simple simplification and 4=2^{2}  and 6=2*3</em>

5. 2^{5/2} \sqrt{3} \sqrt{2} \sqrt{3} *2\sqrt{2} \sqrt{3} *2^{2}\sqrt{2} \sqrt{3}*4\sqrt{3}

<em>collecting the roots on one side and applying exponent rule</em>

6. \sqrt{3} \sqrt{3}\sqrt{3} \sqrt{3}\sqrt{2} \sqrt{2}\sqrt{2} *2^{5/2+1+2+2} \sqrt{3}

<em>Applying exponents rule on all \sqrt{3} and \sqrt{2}</em>

<em>7. 2^{1/2+1/2+1/2} *2^{5/2+1+2+2}*3^{1/2+1/2+1/2+1/2+1/2}</em>

<em>combining all powers of 2</em>

8. 2^{1/2+1/2+1/2+5/2+1+2+2}*3^{1/2+1/2+1/2+1/2+1/2}

<em>Simplifying</em>

9. 2^{9} *3^{2}\sqrt{3}

10. 512*9\sqrt{3}

11. 4608\sqrt{3}

8 0
3 years ago
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