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vovikov84 [41]
2 years ago
15

Find (f•g)(2) given f(x)=x-3 and g(x)=x+3 Please show your work

Mathematics
1 answer:
marissa [1.9K]2 years ago
7 0

Answer:

f.g(x) = f(g(x))\\f(g(2)) = ?\\\\g(x) = x+3 \\g(2) = 5\\\\f(x) = x-3\\f(g(2)) = f(5) = 2=f.g(2)

Step-by-step explanation:

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A shop has 4 fruits (bananas, oranges, apples, and grapes) and 2 dips (peanut butter and caramel).
makvit [3.9K]

Answer:

total fruits and dips=4+2=6

P(select banana&peanut butter)=banana+peanut /total fruits and d

Step-by-step explanation:

total fruits and dips=4+2=6

P(select banana&peanut)=banana+peanut/total fruits and dips

P(select banana&peanut)=1+1=2/6

=1/3

5 0
2 years ago
Please answer and thank you
Aleksandr [31]
5.945 is the answer, can i get brainliest
5 0
3 years ago
Read 2 more answers
Difference between compounded annually, monthly and daily.
zvonat [6]

Answer:

With monthly compounding, the bank will calculate interest on your account just once per month. It will not update your balance on a daily basis when it calculates how much interest it owes you. Assuming that the APR is the same, accounts with monthly compounding offer a lower APY than accounts with daily compounding.

5 0
2 years ago
The difference of the square of a number and 24 is equal to 5 times that number. Find the negative solution.
vampirchik [111]

Answer:

the "negative solution" is -3

Step-by-step explanation:

Represent the number by n.

Then n^2 - 24 = 5n

We rewrite this in standard quadratic form:

n^2 - 5n - 24 = 0

This factors as follows;  (n + 3)(n - 8) = 0

The roots are n = -3 and n = 8.  Thus, the "negative solution" is -3

8 0
3 years ago
Use complete sentences to describe the process in proving that the solution of an inequality is valid.
saul85 [17]
≥The solution of an inequality is an interval, i.e. a range.

To prove that the interval found as solution, you must consider several cases.

1) In the case that the ineguailty is ≥ or ≤, first use the limits of the interval to prove they are valid solutions. This is, replace the limit values, one at a time, and verifiy the inequality.

2) If the sign is ≥ or > use a value to the right of the limit value to show that the values to the right are solution, and use a value to the left to show that they are not solution.

3) If the sign is ≤ or <, use a value to the left of the limit value to show that it is a solution and a value to the right of the limit value to show that it is not a solution. 
7 0
3 years ago
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