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marishachu [46]

2 years ago

13

Which methods could you use to calculate th y-coordinate of the midpoint of a vertical line segment with ndpoints at (0,0) and (

0,-12)? check all that apply
A. divide -12 by 2
B. count by hand
C. divide 12 by 2
D. Add the endpoints

1 answer:

Damm [24]2 years ago

3 0

Answer: B count by hand
Explaining:

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Simplify the number into simplest radical form. Use the factor tree to help determine the factors. StartRoot 96 Endroot StartRoo

NikAS [45]

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. $4608\sqrt{3}$

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This is an improper. Perhaps you can fix it, so that I can assist you with it? I apologise.

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