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2 years ago

For what value of the domain {–3, –1, 0, 2, 5} does f(x) = g(x) When f(x) = 3x – 4 and g(x) = x2 – 2

1 answer:

7 0

F(x) = g(x)
3(2) - 4 = (2)² - 2
6 - 4 = 4 - 2
2 = 2
I just substituted all the values in the domain to see which one would work. That's not really the quickest way but that's the only way I know

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4) Which pair of triangles is congruent by Angle - Angle- Side?

andrey2020 [161]

Answer:

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Step-by-step explanation: Sorry if not correct :(

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2 years ago

F (5^4)m = 512, what is the value of m? 2 3 5 8

MatroZZZ [7]

Answer:

questions is wrong??

Step-by-step explanation:

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2 years ago

If f(x) = 2x^2 + 2, find f(5)

zlopas [31]

Well, following the order of PEMDAS, I got choice B. 52

For instance, when you plug in 5 for x, you get F(5)=2(5)^2+2.

Moreover, following PEMDAS, you're supposed to solve what's inside the parenthesis, but since there is no operation going on inside the parenthesis, then you simple move on to the exponent.

In this case, you square the number 5, which gives you F(5)=2(25)+2

After that, you Multiply (letter M in PEMDAS). This results in F(5)=50+2.

Finally, you add them, which results in F=52.

By the way, I noticed a mistake in your work. When multiplying 2 by 5, the answer is 10, not 20.

Anyway, hope this helped! :-)

8 0

2 years ago

Make r the subject and find the value

nalin [4]

$~~~~~~~~\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac 1{R_2}\\\\\\\implies \dfrac 1{R_2} = \dfrac{1}{R } - \dfrac 1{R_1}\\\\\\\implies \dfrac 1{R_2} = \dfrac{R_1 -R}{RR_1}\\\\\\\implies R_2 = \dfrac{RR_1}{R_1-R}\\\\\\\text{When}~ R= 24~ \Omega ~ \text{and}~ R_1 = 30~ \Omega,\\\\R_2 = \dfrac{24 \times 30}{30-24}~\Omega\\\\\\~~~~~=\dfrac{720}{6}~ \Omega\\\\\\~~~~= 120~ \Omega$

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1 year ago

Can u solve this for me please

zloy xaker [14]

Answer:

-7/32

Step-by-step explanation:

Multiply,Simplify

5 0

2 years ago

Read 2 more answers