SLOPE INTERCEPT
13x − 11y = −12
y =13/11x +12/11
h(x) = 3x + 3
g(x) = −4x + 1
Find (h + g)(10)
−6
Log4 1/64
-3
x2 − 16x + 63
(x − 9)(x − 7)
If you deposit $4000 into an account paying 6% annual interest compounded quarterly, how much money will be in the account after 5 years?
After 5 years there will be $5387.42 in the account.
What is the equation of the line parallel to y=3x+2 and goes through (2,−4)?
y=3x-10
g(a) = 3a + 2
f (a) = 2a − 4
Find (g/f )(3)
11/2
Approximate to the nearest thousand: log7 84
2.276
7x2 − 45x − 28
(7x + 4)(x − 7)
If you deposit $6500 into an account paying 8% annual interest compounded monthly, how much money will be in the account after 7 years?
After 7 years there will be $11358.24 in the account.
through: (3, 5), slope =5/3
b = 0
g(a) = 2a − 1
h(a) = 3a − 3
Find (g ⋅ h)(−4)
135
Rewrite in exponential form: 9-2 = 1/81
log9 1/81 = −2
28n4 + 16n3 − 80n2
4n2 (7n − 10)(n + 2)
How much money would you need to deposit today at 9% annual interest compounded monthly to have $12000 in the account after 6 years?
You would need to deposit $7007.08 to have $12000 in 6 years.
through: (−2, 4), parallel to y = −5/2x + 5
b=-1
g(a) = 2a + 2
h(a) = −2a − 5
Find (g*h)(−4 + a)
−4a + 8
Rewrite in logarithmic form: ( 1/5)x = y
log 1/5y = x
6n2 − 18n − 18 = 6
{4, −1}
If you deposit $8000 into an account paying 7% annual interest compounded quarterly, how long until there is $12400 in the account?
It will take approximately 6.3 years for the account to go from $8000 to $12400.
through: (2, 5), slope = undefined
x = 2
g(n) = 2n + 3
h(n) = n − 1
Find (g*h)(n)
2n + 1
log (4 p − 2) = log (−5 p + 5)
7/9
x3-x2-12x=0
(0,4,-3)
At 3% annual interest compounded monthly, how long will it take to double your money?
At 3% annual interest it will take approximately 23.1 years to double your money