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ra1l [238]
3 years ago
10

I have to factor polynomials by grouping

Mathematics
1 answer:
LUCKY_DIMON [66]3 years ago
6 0

Answer:

Step-by-step explanation:

Factor 6f3−8f2+20−15f

6f^3−8f^3−15f

=(3f−4)(2f2−5)

Answer:

(3f−4)(2f^2−5)

Find the group you want to factor, such as in the first parenthesis, you remove 3f from the whole equation, and end up with 3f-4. You take the end value, and group it with -5. You will then end up with 2f^2 -5.

Hope this Helps!

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Cerrena [4.2K]

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Answer B

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The area of a rectangular piece of paper is 32 square inches. The perimeter is 24 inches.
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What is the volume of the pyramid?
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The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. He wants to first deter
VMariaS [17]

Answer:

He must survey 123 adults.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

Assume that a recent survey suggests that about 87​% of adults have heard of the brand.

This means that \pi = 0.87

90% confidence level

So \alpha = 0.1, z is the value of Z that has a p-value of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

How many adults must he survey in order to be 90​% confident that his estimate is within five percentage points of the true population​ percentage?

This is n for which M = 0.05. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.05 = 1.645\sqrt{\frac{0.87*0.13}{n}}

0.05\sqrt{n} = 1.645\sqrt{0.87*0.13}

\sqrt{n} = \frac{1.645\sqrt{0.87*0.13}}{0.05}

(\sqrt{n})^2 = (\frac{1.645\sqrt{0.87*0.13}}{0.05})^2

n = 122.4

Rounding up:

He must survey 123 adults.

6 0
3 years ago
Which statements are true about the graph of the function f(x) = x2 – 8x + 5? Check all that apply.
statuscvo [17]

Answer:

A, D, E are true

Step-by-step explanation:

You have to complete the square to prove A.  Do this by first setting the function equal to 0, then moving the 5 to the other side.

x^2-8x=-5

Now we can complete the square.  Take half the linear term, square it, and add it to both sides.  Our linear term is 8 (from the -8x).  Half of 8 is 4, and 4 squared is 16.  So we add 16 to both sides.

(x^2-8x+16)=-5+16

We will do the addition on the right, no big deal.  On the left, however, what we have done in the process of completing the square is to create a perfect square binomial, which gives us the h coordinate of the vertex.  We will rewrite with that perfect square on the left and the addition done on the right,

(x-4)^2=11

Now we will move the 11 back over, which gives us the k coordinate of the vertex.

(x-4)^2-11=y

From this you can see that A is correct.

Also we can see that the vertex of this parabola is (4, -11), which is why B is NOT correct.

The axis of symmetry is also found in the h value.  This is, by definition, a positive x-squared parabola (opens upwards), so its axis of symmetry will be an "x = " equation.  In the case of this type of parabola, that "x = " will always be equal to the h value.  So the axis of symmetry is

x = 4, which is why C is NOT correct, either.

We can find the y-intercept of the function by going back to the standard form of the parabola (NOT the vertex form we found by completing the square) and sub in a 0 for x.  When we do that, and then solve for y, we find that when x = 0, y = 5.  So the y-intercept is (0, 5).

From this you can see that D is also correct.

To determine if the parabola has real solutions (meaning it will go through the x-axis twice), you can plug it into the quadratic formula to find these values of x.  I just plugged the formula into my graphing calculator and graphed it to see that it did, indeed, go through the x-axis twice.  Just so you know, the values of x where the function go through are (.6833752, 0) and (7.3166248, 0).  That's why you need the quadratic formula to find these values.

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