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3 years ago

rewrite f(x)=x^2+8x-20 in the form that would most easily help you identify the zeros of the function

Mathematics

2 answers:

Snowcat [4.5K]3 years ago

8 0

Answer:

f(x) = (x-2) (x+10)

Step-by-step explanation:

f(x)=x^2+8x-20

Factor the right hand side

What 2 numbers multiply to -20 and add to 8

-2 * 10 = -20

-2 + 10 = 8

f(x) = (x-2) (x+10)

Then we can use the zero product property to help us to find the zero's

kap26 [50]3 years ago

6 0

Answer:

B.f(x)=(x+6)(x-2)

Step-by-step explanation:

apexs

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Step-by-step explanation:

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2 years ago

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The expression below is the factorization of what trinomial-1(x+5)(x+6)

Nadya [2.5K]

Answer:

-x^2 - 11x -30

Step-by-step explanation:

Solve using foiling. Ignore the -1 to begin with and just look at the part in parenthesis. Do x from the first parenthesis times the stuff in the second parenthesis.

ie: x(x) and x(6)

ie: x^2 + 6x

Then do the 5 times the things in the second parenthesis.

ie: 5(x) and 5(6)

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Then add what you got from multiplying the first value to what you got from multiplying the second.

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This is a trinomial because it has three different variables. Now change all the signs to negative because of the -1 out front.

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2 years ago

Mathematics 13 don’t understand

UNO [17]

Answer:

1/6^3

Step-by-step explanation:

The applicable rules of exponents are ...

a^-b = 1/a^b

(a^b)(a^c) = a^(b+c)

Your expression can be simplified as follows:

$\dfrac{6^{-5}}{6^{-2}}=\dfrac{1}{(6^{-2})(6^5)}=\dfrac{1}{6^{-2+5}}=\boxed{\dfrac{1}{6^3}}$

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<em>Additional comment</em>

If you think of an exponent as signifying repeated multiplication, the rules of exponents may be easier to remember. The exponent tells you how many times the base is a factor in the product.

Consider multiplication:

$(x\cdot x\cdot x)\cdot(x\cdot x)=x^3\cdot x^2=x^{3+2}=x^5\\\\(x\cdot x\cdot x)\cdot(x\cdot x\cdot x)=(x^3)^2=x^{3\cdot2}=x^6$

Consider division:

$\dfrac{x\cdot x\cdot x}{x\cdot x}=x\quad\Longleftrightarrow\quad\dfrac{x^3}{x^2}=x^{3-2}=x^1\\\\\dfrac{x\cdot x}{x\cdot x\cdot x}=\dfrac{1}{x}\quad\Longleftrightarrow\quad\dfrac{x^2}{x^3}=x^{2-3}=x^{-1}$

This may help you see that a positive exponent in the denominator is equivalent to a negative exponent in the numerator (and vice versa).

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3 years ago

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