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ss7ja [257]
3 years ago
5

If f(x) = |x| + 9 and g(x) = –6, which describes the range of (f + g)(x)?

Mathematics
1 answer:
tangare [24]3 years ago
8 0

Answer:

The range of the (f+g)(x) is all real number or ]-\infty,\infty[

Step-by-step explanation:

Let's difne the addition of two funtion as:

(f+g)(x)=f(x)+g(x)

(f+g)(x)=|x|+9-6

(f+g)(x)=|x|+3

The range of the (f+g)(x) is all real number or ]-\infty,\infty[.

I hope it helps you!

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Molly buys 2 boxes of crackers. Each box has 4 packages of crackers in it. Molly
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Need answer quickly! thank you in advance!
anyanavicka [17]

Answer:

b)(b²-a²)

Step-by-step explanation:

a cotθ + b cosecθ =p

b cotθ + a cosecθ =q

Now,

p²- q²

=(a cotθ + b cosecθ)² - (b cotθ + a cosecθ)²     [a²-b²=(a+b)(a-b)]

=(acotθ+bcosecθ + bcotθ+ acosecθ) (a cotθ + bcosecθ -bcotθ-acosecθ)

={a(cotθ+cosecθ)+b(cotθ+cosecθ)} {a (cotθ-cosecθ)+b (cosecθ-cotθ)}

={a(cotθ+cosecθ)+b(cotθ+cosecθ)} [a (cotθ-cosecθ) + {- b (cotθ-cosecθ)} ]

={a(cotθ+cosecθ)+b(cotθ+cosecθ)} {a (cotθ-cosecθ) - b (cotθ-cosecθ)}

={(cotθ+cosecθ)(a+b)} {(cotθ-cosecθ) (a-b)}

=(cotθ+cosecθ) (a+b) (cotθ-cosecθ) (a-b)

=(cotθ+cosecθ) (cotθ-cosecθ) (a+b) (a-b)        

= (cot²θ-cosec²θ) (a²-b²)                                 [(a+b) (a-b)= (a²-b²)]

= -1 . (a²-b²)                               [ 1+cot²θ=cosec²θ ; ∴cot²θ-cosec²θ=-1]

=(b²-a²)

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