Let f(x) = x and g(x) = 7x<br> Find fg()

The expanded form of the whole number 470,052 is

Contact [7]

The answer would be 400,000; 70,000; 50; 2.
Hope this helps!

5 0

2 years ago

Please *atleast* answer one of them!

Julli [10]

Answer and Step-by-step explanation:

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1. Let's write out the equation to subtract.

7t - 2u - 3v - (t - 3v)

Distribute the negative to the t and -3v.

7t - 2u - 3v - t + 3v (The negatives cancel out)

Now simplify by combining like terms.

6t - 2u

This is the answer because the 3v and -3v cancel out.

2. I don't really understand what this is saying. Is there answer choices for this? But what I think its saying is that the lift has a constant of 2.

3. To find out the amount of terms, we would simplify the equation.

2x + 3y - 5x + yz - x

-4x + 3y + yx

Here, we can see that we have 3 terms in this expression.

-4x is the first term, +3y is the second term, and +yx is the third term.

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#teamtrees #WAP (Water And Plant)

8 0

2 years ago

Which is more, 285 inches or 8 yards? 285 inches 8 yards neither; they are equal submit?

Dima020 [189]

285 inches = 7.91667 yards

so if you round up they are equal but if you don't they are not equal

7 0

3 years ago

) An instructor gives his class a set of 18 problems with the information that the next quiz will consist of a random selection

RSB [31]

Answer:

The probability the he or she will answer correctly is 1.5%

Step-by-step explanation:

In all, there are 18 problems. In this question, the order of which the problems are sorted for the quiz makes no difference. For example, if the question A of the quiz is P1 and question B P2, and question A P2 and question B P1, it is the same thing.

There are 18 problems and 9 are going to be selected. So, there is going to be a combination of 9 elements from a set of 18 elements.

A combination of n elements from a set of m objects has the following formula:

$C_{(m,n)} = \frac{m!}{n!(m-n)!}$

In this question, m = 18, n = 9. So the total number of possibilities is:

$T_{p} = C_{(18,9)} = \frac{18!}{9!(18-9)!} = 48620$

Now we have to calculate the number of desired outcomes. This number is a combination of 9 elements from a set of 13 elements(13 is the number of problems that the student has figured out how to do).

Now, m = 13, n = 9. The number of desired possibilities is:

$D_{p} = C_{(13,9)} = \frac{13!}{9!(13-9)!} = 715$