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gayaneshka [121]
3 years ago
14

Use the graph of the polynomial function to find the factored form of the related polynomial. Assume it has no constant factor.

Mathematics
1 answer:
Tpy6a [65]3 years ago
8 0

Answer: C

Step-by-step explanation: If you turn all of them into a x^2 equation yk. A and D would be on the left. While B and C on the right, but B is exactly touching the y-axis when placed on the graph. And C isn’t... It matches perfectly as the one shown on the graph shown.

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Please help 30 POINTSSSSS
sveta [45]

Answer:

-16

Step-by-step explanation:

0.44×10^-15

4.4×10^-16

4 0
2 years ago
I need the answer to this asap, plzz
ira [324]

Answer:

\mathrm{D.\:}\sqrt{126}

Step-by-step explanation:

All three triangles in the figure are similar. Therefore, we can set up the following proportion:

\frac{18}{m}=\frac{m}{7}

Cross-multiply to solve:

m^2=18\cdot 7 = 126, \\\\m=\fbox{$\sqrt{126}$}.

3 0
3 years ago
Adjacent angles have no common interior points
qwelly [4]
<span>They share a vertex and side, but do not overlap 


Hope this helps</span>
4 0
4 years ago
Solve for y.<br> r/3-2/y=s/5
nordsb [41]

Answer:

y = 2 / (r/3 - s/5)

Step-by-step explanation:

r/3 - 2/y = s/5

add 2/y to both sides

r/3 = s/5 + 2/y

Subtract s/5 from both sides

r/3 - s/5 = 2/y

multiply both sides by y

y(r/3 - s/5) = 2

Divide both sides by r/3 - s/5

y = 2 / (r/3 - s/5)

8 0
3 years ago
Please find the derivative of that function...
vodka [1.7K]
You could apply a clever Algebra trick to avoid using the quotient rule,

\rm y=\dfrac{log(x)-1+1}{log(x)-1}=\dfrac{log(x)-1}{log(x)-1}+\dfrac{1}{log(x)-1}

\rm y=1+\dfrac{1}{log(x)-1}

and apply power rule into chain rule from that point.
But this problem was likely designed to teach you about quotient rule so let's do it that way.

Let's start by "setting up" our quotient rule:

\rm y'=\dfrac{(log x)'(log x-1)-log x(log x-1)'}{(log x-1)^2}

If this log notation is not intended to be natural log, then we'll have a little bit of extra work. Our change of base identity allows us to rewrite log base 10 in terms of the natural log,

\rm log(x)=\dfrac{ln x}{ln10}

so let's apply this to our problem,

\rm y'=\dfrac{\left(\frac{ln x}{ln 10}\right)'(log x-1)-log x\left(\dfrac{ln x}{ln 10}-1\right)'}{(log x-1)^2}

Derivative of ln(x) gives us 1/x in each case,

\rm y'=\dfrac{\left(\frac{1}{x ln 10}\right)(log x-1)-log x\left(\dfrac{1}{x ln 10}\right)}{(log x-1)^2}

Factor the 1/(x ln10) out of each term in the numerator,

\rm y'=\left(\dfrac{1}{x ln 10}\right)\dfrac{log x-1-log x}{(log x-1)^2}

and combine like-terms,

\rm y'=\dfrac{-1}{x(log x-1)^2ln 10}

Lemme know if you're confused with any of the steps.
6 0
4 years ago
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