Let f(x) = 5x – 3 and g(x) = x² + 2x + 5. Evaluate the following. -
-
a. f(2)=
b. f( – 3) =
c. g( – 2) =
d. g(1) =
e. f(a + 1) =
Preview
f. g(a + 1) =

a. 7

b. -18

c. 5

d. 8

e. 5a + 2

f. a² + 4a + 8

Step-by-step explanation:

f(x) = 5x - 3 and g(x) = x² + 2x + 5

for each, we simply plug in what's inside of the parenthesis wherever we see x

f(2) = 5(2) - 3

multiply 5 and 2

f(2) = 10 - 3 = 7

f(-3) = 5(-3) - 3

multiply 5 and -3

f(-3) = -15 - 3 = -18

g(-2) = (-2)² + 2(-2) + 5

evaluate exponent

g(-2) = 4 + 2(-2) + 5

multiply 2 and -2

g(-2) = 4 - 4 + 5 = 5

( the 4's cancel out and we're left with 5 )

g(1) = (1)² + 2(1) + 5

evaluate exponent

g(1) = 1 + 2(1) + 5

multiply 2 and 1

g(1) = 1 + 2 + 5 = 8

f(a+1) = 5(a + 1) - 3

distribute the 5

f(a+1) = 5a + 5 - 3

combine like terms

f(a+1) = 5a + 2

g(a+1) = (a+1)² + 2(a+1) + 5

expand exponent

g(a+1) = (a+1)(a+1) + 2(a+1) + 5

evaluate (a+1)(a+1) using FOIL

g(a+1) = a² + 2a + 1 + 2(a+1) + 5

distribute the 2

g(a+1) = a² + 2a + 1 + 2a + 2 + 5

combine like terms

g(a+1) = a² + 4a + 8

whole numbers.

step-by-step explanation:

we are given the average monthly rainfall (in millimeters) model r(m) for the years 1900-2009 as a function of the month number m.

as we can see m is taken for the number of months for years 1900-2009.

we know that number of months can't be a negative number or decimal number or a fraction.

so, the number of months could be only positive whole numbers.

therefore, whole numbers is more appropriate for the domain of r.

alright well the answers to your question is

2xy/9 hope this have a nice day : )

step-by-step explanation: