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3 years ago

What are the zeros of f(x) = x2 – 8x + 16?

1 answer:

5 0

<span>What are the zeros of f(x) = x2 – 8x + 16?
</span>
let the function f(x) = 0
so

x^2 – 8x + 16 = 0
(x - 4)^2 = 0
x - 4 = 0
x = 4

hope it helps

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How do you solve these equations? I don't want you to answer all of them, just tell me how to solve each type of equation on the

miss Akunina [59]

Answer:

8. Identify the common denominator; express each fraction using that denominator; combine the numerators of those rewritten fractions and express the result over the common denominator. Factor out any common factors from numerator and denominator in your result. (It's exactly the same set of instructions that apply for completely numerical fractions.)

9. As with numerical fractions, multiply the numerator by the inverse of the denominator; cancel common factors from numerator and denominator.

10. The method often recommended is to multiply the equation by a common denominator to eliminate the fractions. Then solve in the usual way. Check all answers. If one of the answers makes your multiplier (common denominator) be zero, it is extraneous. (10a cannot have extraneous solutions; 10b might)

Step-by-step explanation:

For a couple of these, it is helpful to remember that (a-b) = -(b-a).

$\dfrac{5}{x+2}+\dfrac{25-x}{x^2-3x-10}=\dfrac{5(x-5)}{(x+2)(x-5)}+\dfrac{25-x}{(x+2)(x-5)}\\\\=\dfrac{5x-25+25-x}{(x+2)(x-5)}=\dfrac{4x}{x^2-3x-10}$

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$\displaystyle\frac{\left(\frac{x}{x-2}\right)}{\left(\frac{2x}{2-x}\right)}=\frac{x}{x-2}\cdot\frac{-(x-2)}{2x}=\frac{-x(x-2)}{2x(x-2)}=-\frac{1}{2}$

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$\dfrac{3}{x-1}+\dfrac{6}{x^2-3x+2}=2\\\\\dfrac{3(x-2)}{(x-1)(x-2)}+\dfrac{6}{(x-1)(x-2)}=\dfrac{2(x-1)(x-2)}{(x-1)(x-2)}\\\\3x-6+6=2(x^2-3x+2) \qquad\text{multiply by the denominator}\\\\2x^2-9x+4=0 \qquad\text{subtract 3x}\\\\(2x-1)(x-4)=0 \qquad\text{factor; x=1/2, x=4}$

Neither solution makes any denominator be zero, so both are good solutions.

8 0

3 years ago

Make 478 three hundred times as large

dexar [7]

478x2= 956
956x2= 1912

I hope this is right :)

7 0

3 years ago

Read 2 more answers

I really need help with this question

SOVA2 [1]

In the matrix, the first row is exactly twice the second. This implies that the determinant is zero.

If the matrix system has determinant zero, the system has either no solutions or infinite solutions.

However, just like in the matrix, in the constant vector the first term is twice the second. This means that the system has infinite solutions.

In fact, if you write this system in expanded form, you can see that it generates two equations, the first being twice the second:

$\begin{cases}2x+4y=6\\x+2y=3\end{cases}$

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3 years ago

Help me plz im stuck

viva [34]

Answer:

A) Fill the blanks, Fail, 0.15, Success, 244, 0.61

B) 240 times

Step by-step explanation:

A) Void , 96, 0.24

Fail, 60, it should be 0.15 because 60/400

Success, 400 - 96- 60 or 244 and the relative frequency is 0.61.

There is a total of 1000. If the relative frequency is 0.24, multiply 0.24 by 1000 to get the amount of times YOU would expect to be VOID.

Or in other words , 240 times

By proving it add all up

0.24 x 1000 + 0.76 x 1000 = 1000

<h3><u><em>Hope this helps!!!</em></u></h3><h3><u><em>PLEASE GIVE BRAINLIEST!!!</em></u></h3><h3><u><em>APPRECIATE IT!!!</em></u></h3><h3><u><em>AND AS ALWAYS,</em></u></h3><h3><u><em>SIMPLIFY BANANAS : )</em></u></h3>

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3 years ago

What is the value of 4x^-3y^3 for x = 3 and y=-1?

il63 [147K]

Answer:

− 23 /27

its a fraction

Step-by-step explanation:

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3 years ago