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Kitty [74]
2 years ago
5

Complete the following statement.

Mathematics
1 answer:
Afina-wow [57]2 years ago
4 0

Answer:

x=2; the extraneous root x=42.

All the details are in the attached picture, the answer is marked with red colour.

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Evaluate each function.<br> 1) f(x) = x2 + 4x; Find f(-7)
katrin [286]

Answer:

The answer is -77

Step-by-step explanation:

Ok, so assuming by x2 you mean x squared, I will solve this. So basically when you have a function, f(-7) would mean that you would have to replace all the x's in the equation with -7. So let's write that out. that would be f(-7) = -7^2 + (-7*4). So now according to PEMDAS, you would solve the exponent first, and -7^2 is equal to -49, because when you solve it you would do -(7^2), which is -(49), which is then -49. So now you have f(-7)= -49+ (4*-7). Solving for (4*-7), you get -28. This leaves you with -49 + (-28), which is -49 - 28. Simplifying that, you get the answer, which is -77.

5 0
3 years ago
Find the product of this please thank you
iVinArrow [24]
You combine the like terms to get
2s^2y^2-4a^2sy+2a^4
7 0
3 years ago
What is the square root of 0.0225
ohaa [14]

Answer:

0.15

Step-by-step explanation:

i goog.led it its probably right

6 0
3 years ago
Read 2 more answers
Which of the following is the simplified form of fifth root of x times the fifth root of x times the fifth root of x times the f
Juliette [100K]

Answer:

\large\boxed{x^\frac{4}{5}}

Step-by-step explanation:

\sqrt[n]{a}=a^\frac{1}{n}\Rightarrow\sqrt[5]{x}=x^\frac{1}{5}\\\\\sqrt[5]{x}\cdot\sqrt[5]{x}\cdot\sqrt[5]{x}\cdot\sqrt[5]{x}=x^\frac{1}{5}\cdot x^\frac{1}{5}\cdot x^\frac{1}{5}\cdot x^\frac{1}{5}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=x^{\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}}=x^\frac{4}{5}

4 0
3 years ago
Read 2 more answers
What is the quadratic function f(x)=x^2+6x-2 In vertex form?
Yuliya22 [10]

Answer: Option D

f(x)=(x+3)^2 -k

Step-by-step explanation:

For a quadratic function of the form

ax ^ 2 + bx + c

The x coordinate of the vertice is:

x =-\frac{b}{2a}

In this case the function is:

f(x)=x^2+6x-2\\\\

So

a=1\\b=6\\c=-2

The x coordinate of the vertice is:

x=-\frac{6}{2*1}\\\\x=-3

The y coordinate of the vertice is:

f(-3) = (-3)^2 +6(-3) -2\\\\f(-3)=-11

The vertice is: (-3, -11)

The form e vertice for a quadratic equation is:

f(x)=(x-h)^2 +k

Where

the x coordinate of the vertice is h and the y coordinate of the vertice is k.

Then h=-3 and k =-11

Finally the equation f(x)=x^2+6x-2\\\\ in vertex form is:

f(x)=(x+3)^2 -k

6 0
3 years ago
Read 2 more answers
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