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>
Yes. When the function f(x) = x3 – 75x + 250 is divided by x + 10, the remainder is zero. Therefore, x + 10 is a factor of f(x) = x3 – 75x + 250.
According to the remainder theorem when f(x) is divided by (x+a) the remainder is f(-a).
In this case,
f(x)=x^3-75x+250
(x+a)=(x+10)
Therefore, the remainder f(-a)=f(-10)
=x^3-75x+250
=(-10)^3-(75*-10)+250
=-1000+750+250
=1000-1000
=0.
The remainder is 0. So, (x+10) is a factor of x^3-75x+250.
Answer:
question does not seem complete, from what i know it sounds like he should add up the test scores and divide it by five, that will give him his average which should be his grade, please update question if possible
i hope this helped ;)
Answer:
Step-by-step explanation:
here
