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TEA [102]
3 years ago
5

What are the zeros for the function f(x)=x^3-4x

Mathematics
1 answer:
Harman [31]3 years ago
4 0

Answer:

x = 0

x = 2

x = -2

Step-by-step explanation:

f(x) = x^{3} - 4x

Factor

f(x) = x ( x^{2} - 4)

f(x) = x ( ( x - 2 ) ( x + 2 ) )

zeros;

x = 0

x = 2

x = -2

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Margarita [4]

Answer:

It comes down to what the equals sign means. It means,  that the two things mentioned on either side of the equation are the same thing. If you do something to the left side, you have presumably changed it.

Step-by-step explanation:

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8 0
3 years ago
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Eric has an estimated income of $2,270 per month. After budgeting his expenses he has found that he has about $836 remaining. If
vladimir1956 [14]

1,434÷2,270=0.63×100=63%





5 0
4 years ago
Y=|x| (What is the range of this function?)
olasank [31]

Answer:

Any number both positive or negative.

y = {all numbers}

Step-by-step explanation:

5 0
3 years ago
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❊ Simplify :
DiKsa [7]

Answer:

See Below.

Step-by-step explanation:

Problem 1)

We want to simplify:

\displaystyle \frac{a+2}{a^2+a-2}+\frac{3}{a^2-1}

First, let's factor the denominators of each term. For the second term, we can use the difference of two squares. Hence:

\displaystyle =\frac{a+2}{(a+2)(a-1)}+\frac{3}{(a+1)(a-1)}

Now, create a common denominator. To do this, we can multiply the first term by (<em>a</em> + 1) and the second term by (<em>a</em> + 2). Hence:

\displaystyle =\frac{(a+2)(a+1)}{(a+2)(a-1)(a+1)}+\frac{3(a+2)}{(a+2)(a-1)(a+1)}

Add the fractions:

\displaystyle =\frac{(a+2)(a+1)+3(a+2)}{(a+2)(a-1)(a+1)}

Factor:

\displaystyle =\frac{(a+2)((a+1)+3)}{(a+2)(a-1)(a+1)}

Simplify:

\displaystyle =\frac{a+4}{(a-1)(a+1)}

We can expand. Therefore:

\displaystyle =\frac{a+4}{a^2-1}

Problem 2)

We want to simplify:

\displaystyle \frac{1}{(a-b)(b-c)}+\frac{1}{(c-b)(a-c)}

Again, let's create a common denominator. First, let's factor out a negative from the second term:

\displaystyle \begin{aligned} \displaystyle &= \frac{1}{(a-b)(b-c)}+\frac{1}{(-(b-c))(a-c)}\\\\&=\displaystyle \frac{1}{(a-b)(b-c)}-\frac{1}{(b-c)(a-c)}\\\end{aligned}

Now to create a common denominator, we can multiply the first term by (<em>a</em> - <em>c</em>) and the second term by (<em>a</em> - <em>b</em>). Hence:

\displaystyle =\frac{(a-c)}{(a-b)(b-c)(a-c)}-\frac{(a-b)}{(a-b)(b-c)(a-c)}

Subtract the fractions:

\displaystyle =\frac{(a-c)-(a-b)}{(a-b)(b-c)(a-c)}

Distribute and simplify:

\displaystyle =\frac{a-c-a+b}{(a-b)(b-c)(a-c)}=\frac{b-c}{(a-b)(b-c)(a-c)}

Cancel. Hence:

\displaystyle =\frac{1}{(a-b)(a-c)}

4 0
3 years ago
Margie says the set of numbers, 0.1, -1, -2 1/4, 3, are in order from least to greatest. What is her error?
ANEK [815]

Her error was that she thought 0.1 and 1/4 were negatives but they are still positive.

She also thought the larger the negative the greatest one. The smaller negative is the greatest because it's bellow 0.

To solve this you would first need to turn 1/4 into 0.25

It would start out with -2 cause that is the least. Then -1. Now its back to positive so the least to greatest would be normal.

So the real answer would be -2, -1, 0.1, 1/4

7 0
3 years ago
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