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Bond [772]
3 years ago
14

I have a Algebra 2 question f(x) = 5x +4 and g(x) = x^2 - 5 find g(f(4))

Mathematics
2 answers:
Alenkasestr [34]3 years ago
7 0

Answer:

g(f(4)) = 571

Step-by-step explanation:

g(f(4))

We need to evaluate the function f at x=4

f(4) = 5*4 +4

f(4) = 20+4

f(4) = 24

Now we have to evaluate the  function g at 24

g(24) = 24^2 -5

             576 -5

           571


g(f(4)) = 571

victus00 [196]3 years ago
5 0

Answer: 571

To get this answer, we first need to compute the value of f(4). This happens when we plug x = 4 into f(x)

f(x) = 5x + 4

f(4) = 5*4+4 ... each x replaced with 4

f(4) = 20+4

f(4) = 24

Now replace every 'x' in g(x) with f(4) like so

g(x) = x^2 - 5

g(x) = ( x )^2 - 5

g(f(4)) = ( f(4) )^2 - 5 .... every x replaced with f(4)

g(f(4)) = ( 24 )^2 - 5 ... see note below

g(f(4)) = 576 - 5

g(f(4)) = 571

note: the f(4) on the right hand side of that equation is replaced with 24, because we found earlier that f(4) = 24. In other words, f(4) is the same as 24.

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In-s [12.5K]

Answer:

\mathbf{\iiint_W (x^2+y^2) \ dx \ dy \ dz = \dfrac{2}{15}}

Step-by-step explanation:

Given that:

\iiint_W (x^2+y^2) \ dx \ dy \ dz

where;

the top vertex = (0,0,1) and the  base vertices at (0, 0, 0), (1, 0, 0), (0, 1, 0), and (1, 1, 0)

As such , the region of the bounds of the pyramid is: (0 ≤ x ≤ 1-z, 0 ≤ y ≤ 1-z, 0 ≤ z ≤ 1)

\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \int ^{1-z}_0 \int ^{1-z}_0 (x^2+y^2) \ dx \ dy \  dz

\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \int ^{1-z}_0 ( \dfrac{(1-z)^3}{3}+ (1-z)y^2) dy \ dz

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\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0  \ dz \  ( \dfrac{(1-z)^4}{3}+ \dfrac{(1-z)^4}{3}) \ dz

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\mathbf{\iiint_W (x^2+y^2) \ dx \ dy \ dz = \dfrac{2}{15}}

7 0
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aleksandr82 [10.1K]

Answer:

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Since they are independent events, we can multiply the probabilities together

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4 0
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Answer:

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Lelu [443]
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</span>I hope this helps.
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kondaur [170]
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