D. Tanisha is 6, and Bess is 12
The answer to the problem is A
Simplify each term<span>.</span>
Simplify <span>3log(x)</span><span> by moving </span>3<span> inside the </span>logarithm<span>.
</span><span>log(<span>x^3</span>)+2log(y−1)−5log(x)</span><span>
</span>
Simplify <span>2log(y−1)</span><span> by moving </span>2<span> inside the </span>logarithm<span>.
</span><span>log(<span>x^3</span>)+log((y−1<span>)^2</span>)−5log(x)</span><span>
</span>
Rewrite <span>(y−1<span>)^2</span></span><span> as </span><span><span>(y−1)(y−1)</span>.</span><span>
</span><span>log(<span>x^3</span>)+log((y−1)(y−1))−5log(x)</span><span>
</span>
Expand <span>(y−1)(y−1)</span><span> using the </span>FOIL<span> Method.
</span><span>log(<span>x^3</span>)+log(y(y)+y(−1)−1(y)−1(−1))−5log(x)</span><span>
</span>
Simplify each term<span>.
</span><span>log(<span>x^3</span>)+log(<span>y^2</span>−2y+1)+log(<span>x^<span>−5</span></span>)</span><span>
</span>Remove the negative exponent<span> by rewriting </span><span>x^<span>−5</span></span><span> as </span><span><span>1/<span>x^5</span></span>.</span><span>
</span><span>log(<span>x^3</span>)+log(<span>y^2</span>−2y+1)+log(<span>1/<span>x^5</span></span>)</span><span>
</span>
Combine<span> logs to get </span><span>log(<span>x^3</span>(<span>y^2</span>−2y+1))
</span><span>log(<span>x^3</span>(<span>y^2</span>−2y+1))+log(<span>1/<span>x^5</span></span>)
</span>Combine<span> logs to get </span><span>log(<span><span><span>x^3</span>(<span>y^2</span>−2y+1)/</span><span>x^5</span></span>)</span><span>
</span>log(x^3(y^2−2y+1)/x^5)
Cancel <span>x^3</span><span> in the </span>numerator<span> and </span>denominator<span>.
</span><span>log(<span><span><span>y^2</span>−2y+1/</span><span>x^2</span></span>)</span><span>
</span>Rewrite 1<span> as </span><span><span>1^2</span>.</span>
<span><span>y^2</span>−2y+<span>1^2/</span></span><span>x^2</span>
Factor<span> by </span>perfect square<span> rule.
</span><span>(y−1<span>)^2/</span></span><span>x^2</span>
Replace into larger expression<span>.
</span>
<span>log(<span><span>(y−1<span>)^2/</span></span><span>x^2</span></span>)</span>
The answer is 30.
First off, the numbers are consecutive even numbers. So, the difference between every consecutive number is 2 (numbers go from even to odd to even to odd...). Since the sum of their numbers is 96, I divided that by 3 to get 32. This gives me the median of the three numbers. To find the smallest number, I simply subtracted 2 from the 32.
This may help:
96/3=32
32 + 32 + 32 = 96
(32-2)+(32+0)+(32+2)=96
30+32+34=93
Answer:
81,000
Step-by-step explanation:
go to whatever place it is asking you to round nearest to the if it is 5 or more you round up 4 or less it stays the same.
Saying for this -" 5 or more raise the score, 4 or less let it rest"
^round up^ ^stays the same^
