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>
Your answer is A) (1,1) is the solution to both lines K and J
That's because the solution to a system of equations has to be the solution to both of the lines. Also, both of the lines intersect at the point (1,1).
Answer:
Critical Value
Step-by-step explanation:
The critical value separates the region of rejection from the region of non rejection.
We define critical values as:
- Critical value helps us to give the decision rule for the rejection or acceptance of null hypothesis.
- Critical value separates the acceptance region from the rejection region.
- These are referred to as cut-off values that define regions where the test statistic is unlikely to lie.
- The critical value depends on the kind of test being performed and the significance level.
The correct answer is not up there. its -125a^7
The answer is A is equal to 2
