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F(x) = 2x – 2 g(x)= x² + 3x Find (f.g) (2)

Alex_Xolod [135]

Answer: 18

Hey bud, just as a sidenote, I think you mean f(g(2)) for the question. If this was not what it meant, please take a screenshot of the question.

2^2 + 3*2 = 10

2*10 -2 = 18

6 0

4 years ago

The ratio 2/8 could represent which of the following?

d1i1m1o1n [39]

Answer:

C) 2parts that are one eight each.

Step-by-step explanation:

The reason I chose C) as the correct answer is becasue I evaluated it with the steps below.

A) - 8 parts that are one fourth each.

This cannot be correct because it would surpase the 2/8ths (would be over).

B) - 4 parts that are one-half each.

This cannot be correct because it would surpass the 2/8ths (would be over).

C) - 2parts that are one eight each.

This can be correct because 2 parts have 1/8th. 1/8 + 1/8 = 2/8.

D) - 2parts that are one fourth each.

This cannot be correct because the denominator is not 8, it is 4 meaning it wouldn't fit into the format the question asked for (which was 2/8).

This makes the correct answer C).

I hope this helps you understand and helps you in your question.

8 0

3 years ago

Conditional statement for-the product of two even numbers is even

Nata [24]

If two even numbers are multiplied together, then the product is even

5 0

3 years ago

What is 349 x 206? :((((((((((((

ELEN [110]

Answer:

71894 hope it help

Step-by-step explanation:

8 0

3 years ago

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Which of the following shows the extraneous solution to the logarithmic equation? log Subscript 4 Baseline (x) + log Subscript 4

vekshin1

Given:

The given equation is $\log _{4}(x)+\log _{4}(x-3)=\log _{4}(-7 x+21)$

We need to determine the extraneous solution of the equation.

Solving the equation:

To determine the extraneous solution, we shall first solve the given equation.

Applying the log rule $\log _{c}(a)+\log _{c}(b)=\log _{c}(a b)$ , we get;

$\log _{4}(x(x-3))=\log _{4}(-7 x+21)$

Again applying the log rule, if $\log _{b}(f(x))=\log _{b}(g(x))$ then $f(x)=g(x)$

Thus, we have;

$x(x-3)=-7 x+21$

Simplifying the equation, we get;

$x^2-3x=-7 x+21$

$x^2+4x=21$

$x^2+4x-21=0$

Factoring the equation, we get;

$(x-3)(x+7)=0$

Thus, the solutions are $x=3, x=-7$

Extraneous solutions:

The extraneous solutions are the solutions that does not work in the original equation.

Now, to determine the extraneous solution, let us substitute x = 3 and x = -7 in the original equation.

Thus, we get;

$\log _{4}(3)+\log _{4}(3-3)=\log _{4}(-7 \cdot 3+21)$

$\log _{4}(3)+\log _{4}(0)=\log _{4}(0)$

Since, we know that $\log _{a}(0)$ is undefined.

Thus, we get;

Undefined = Undefined

This is false.

Thus, the solution x = 3 does not work in the original equation.

Hence, x = 3 is an extraneous solution.

Similarly, substituting x = -7, in the original equation. Thus, we get;

$\log _{4}(-7)+\log _{4}(-7-3)=\log _{4}(-7(-7)+21)$

$\log _{4}(-7)+\log _{4}(-10)=\log _{4}(49+21)$

$\log _{4}(-7)+\log _{4}(-10)=\log _{4}(70)$

Simplifying, we get;

Undefined = $\log _{4}(70)$

Undefined = 3.06

This is false.

Thus, the solution x = -7 does not work in the original solution.

Hence, x = -7 is an extraneous solution.

Therefore, the extraneous solutions are x = 3 and x = -7

7 0

3 years ago

Read 2 more answers