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tia_tia [17]
3 years ago
7

Find all of the zeros of the polynomial function below, given that it has one zero at x=1.

Mathematics
1 answer:
babymother [125]3 years ago
3 0

Answer:

10

Step-by-step explanation:

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sweet [91]

I'd love to answer but what's the question

Step-by-step explanation:

please explain

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4 years ago
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Two poles of lengths 10 ft and 15 ft are set up vertically with bases on horizontal ground 12 ft apart. Find the distance betwee
svp [43]

The distance between the tops of the poles is the hypotenuse of the triangle which is equal to the square root of (12^2 + 5^2)

The hypotenuse = square root (144 + 25) = 13 ft

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Abigail and Spencer calculate the slope of the line between the points (3,-1) and (5,4) in different ways. Abigail calculates th
Annette [7]
Abigail and Spencer are correct, Lauren is incorrect.

8 0
2 years ago
How many times can 41 go into 328
Brums [2.3K]
Divide. 328 ÷ 41 = 8
6 0
3 years ago
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Find the distance from the origin to the graph of 7x+9y+11=0
Cerrena [4.2K]
One way to do it is with calculus. The distance between any point (x,y)=\left(x,-\dfrac{7x+11}9\right) on the line to the origin is given by

d(x)=\sqrt{x^2+\left(-\dfrac{7x+11}9\right)^2}=\dfrac{\sqrt{130x^2+154x+121}}9

Now, both d(x) and d(x)^2 attain their respective extrema at the same critical points, so we can work with the latter and apply the derivative test to that.

d(x)^2=\dfrac{130x^2+154x+121}{81}\implies\dfrac{\mathrm dd(x)^2}{\mathrm dx}=\dfrac{260}{81}x+\dfrac{154}{81}

Solving for (d(x)^2)'=0, you find a critical point of x=-\dfrac{77}{130}.

Next, check the concavity of the squared distance to verify that a minimum occurs at this value. If the second derivative is positive, then the critical point is the site of a minimum.

You have

\dfrac{\mathrm d^2d(x)^2}{\mathrm dx^2}=\dfrac{260}{81}>0

so indeed, a minimum occurs at x=-\dfrac{77}{130}.

The minimum distance is then

d\left(-\dfrac{77}{130}\right)=\dfrac{11}{\sqrt{130}}
4 0
4 years ago
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