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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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X+2y=10. x+y=6 SOLVE for the value of x and y.
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</span><span>x+y=6---------Eq.2
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3 years ago
A box contains 8 two-inch screws. Four have a Phillips head and 4 have a slotted head. In how many ways can 4 screws be chosen s
hammer [34]

Answer:

There are 6\cdot 6=36 different ways to choose 4 screws such that 2 have a Phillips head and 2 have a slotted head.

Step-by-step explanation:

If 4 screws must be chosen so that 2 have a Phillips head and 2 have a slotted head, then you have to choose 2 screws with a Phillips head from 4 screws with a Phillips head and 2 screws with a slotted head from 4 screws with a slotted head.

You can choose 2 screws with a Phillips head from 4 screws with a Phillips head in

C^4_2=\dfrac{4!}{2!(4-2)!}=\dfrac{4!}{2!\cdot2!}=\dfrac{1\cdot2\cdot3\cdot4}{1\cdot2\cdot1\cdot2}=6

different ways.

You can choose 2 screws with a slotted head from 4 screws with a slotted head in

C^4_2=\dfrac{4!}{2!(4-2)!}=\dfrac{4!}{2!\cdot2!}=\dfrac{1\cdot2\cdot3\cdot4}{1\cdot2\cdot1\cdot2}=6

different ways.

In total there are 6\cdot 6=36 different ways to choose 4 screws such that 2 have a Phillips head and 2 have a slotted head.

7 0
3 years ago
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