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poizon [28]
4 years ago
7

What are the zeros of the polynomial function f(x) = x3 – 4x2 – 12x

Mathematics
2 answers:
trapecia [35]4 years ago
8 0
X³ - 4x² - 12x
x(x² - 4x - 12)
x(x - 6)(x + 2)
x = 0
x = 6
x = -2
The zeros are 0, 6, and -2
AleksAgata [21]4 years ago
4 0
<span><span>fx</span>=<span><span><span>x3</span>−<span>4<span>x2</span></span></span>−<span>12x</span></span></span>.<span><span><span>‌<span><span>fx/</span>x</span></span>‌</span>=<span><span>‌<span><span><span><span>x3</span>−<span>4<span>x2</span></span></span>−<span>12x/</span></span>x</span></span>‌

</span></span><span>f=<span><span><span>x2</span>−<span>4x</span></span>−<span>12</span></span></span>
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PLEASE HELP ASAP WILL MARK BRAINLIEST!!
Papessa [141]

Answer:

Prime.

Step-by-step explanation:

x^2-7x+8\\

First let's multiply our outer coefficients in this case 1 and 8, so we have 1x8=8.  Now the point of this is to get the factors of 8 and see if we can get that middle term -7.  Factors of 8 are 1x8, 2x4 and that's it.  From there I notice that -8+1 = -7 and so:

x^2-7x+8 = x^2-8x+x+8 = x(x-8)+(x+8)

Notice that the (x-8) and (x+8) are not the same factors, therefore the factors are not any of those answer choices and so it's prime.  You can calculate the factors by using the quadratic formula, however I think that is beyond the scope of this question.

7 0
3 years ago
Andre paid $13 for 3 books. Diego bought 12 books priced at the same rate.
yKpoI14uk [10]
X = $ 52 for the 12 books Diego bought
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3 years ago
6. What’s the answer to this question
Artyom0805 [142]

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C. The number of dog toys

Step-by-step explanation:

3 0
3 years ago
12dogs to 92 cats whats the ratio as a fraction in simplest form
DedPeter [7]

12 dogs to 92 cats.

12/92

Both 12 and 92 are divisible by 4.

12/4=3

92/4=23

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There are 3 dogs for every 23 cats.

4 0
3 years ago
What is a quick and easy way to remember explicit and recursive formulas?
Oliga [24]
I always found derivation to be helpful in remembering. Since this question is tagged as at the middle school level, I assume you've only learned about arithmetic and geometric sequences.

First, remember what these names mean. An arithmetic sequence is a sequence in which consecutive terms are increased by a fixed amount; in other words, it is an additive sequence. If a_n is the nth term in the sequence, then the next term a_{n+1} is a fixed constant (the common difference d) added to the previous term. As a recursive formula, that's

a_{n+1}=a_n+d

This is the part that's probably easier for you to remember. The explicit formula is easily derived from this definition. Since a_{n+1}=a_n+d, this means that a_n=a_{n-1}+d, so you plug this into the recursive formula and end up with 

a_{n+1}=(a_{n-1}+d)+d=a_{n-1}+2d

You can continue in this pattern, since every term in the sequence follows this rule:

a_{n+1}=a_{n-1}+2d
a_{n+1}=(a_{n-2}+d)+2d
a_{n+1}=a_{n-2}+3d
a_{n+1}=(a_{n-3}+d)+3d
a_{n+1}=a_{n-3}+4d

and so on. You start to notice a pattern: the subscript of the earlier term in the sequence (on the right side) and the coefficient of the common difference always add up to n+1. You have, for example, (n-2)+3=n+1 in the third equation above.

Continuing this pattern, you can write the formula in terms of a known number in the sequence, typically the first one a_1. In order for the pattern mentioned above to hold, you would end up with

a_{n+1}=a_1+nd

or, shifting the index by one so that the formula gives the nth term explicitly,

a_n=a_1+(n-1)d

Now, geometric sequences behave similarly, but instead of changing additively, the terms of the sequence are scaled or changed multiplicatively. In other words, there is some fixed common ratio r between terms that scales the next term in the sequence relative to the previous one. As a recursive formula,

a_{n+1}=ra_n

Well, since a_n is just the term after a_{n-1} scaled by r, you can write

a_{n+1}=r(ra_{n-1})=r^2a_{n-1}

Doing this again and again, you'll see a similar pattern emerge:

a_{n+1}=r^2a_{n-1}
a_{n+1}=r^2(ra_{n-2})
a_{n+1}=r^3a_{n-2}
a_{n+1}=r^3(ra_{n-3})
a_{n+1}=r^4a_{n-3}

and so on. Notice that the subscript and the exponent of the common ratio both add up to n+1. For instance, in the third equation, 3+(n-2)=n+1. Extrapolating from this, you can write the explicit rule in terms of the first number in the sequence:

a_{n+1}=r^na_1

or, to give the formula for a_n explicitly,

a_n=r^{n-1}a_1
6 0
3 years ago
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