Solve the inequality. -3 ≤ 2t ≤ 6
2 answers:
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Let x represent amount invested in the higher-yielding account.
We have been given that a man puts twice as much in the lower-yielding account because it is less risky. So amount invested in the lower-yielding account would be .
We are also told that his annual interest is $6600 dollars. We know that annual interest for one year will be principal amount times interest rate.
, where,
I = Amount of interest,
P = Principal amount,
r = Annual interest rate in decimal form,
t = Time in years.
We are told that interest rates are 6% and 10%.
Amount of interest earned from lower-yielding account: .
Amount of interest earned from higher-yielding account: .
Let us solve for x.
Therefore, the man invested $30,000 at 10%.
Amount invested in the lower-yielding account would be .
Therefore, the man invested $60,000 at 6%.
Answer:
Step-by-step explanation:
We are factoring
So:
((2•5^2x^2) + 485x) - 150
Pull like factors :
50x^2 + 485x - 150 = 5 • (10x^2 + 97x - 30)
Factor
10x^2 + 97x - 30
Step-1: Multiply the coefficient of the first term by the constant 10 • -30 = -300
Step-2: Find two factors of -300 whose sum equals the coefficient of the middle term, which is 97.
-300 + 1 = -299
-150 + 2 = -148
-100 + 3 = -97
-75 + 4 = -71
-60 + 5 = -55
-50 + 6 = -44
-30 + 10 = -20
-25 + 12 = -13
-20 + 15 = -5
-15 + 20 = 5
-12 + 25 = 13
-10 + 30 = 20
-6 + 50 = 44
-5 + 60 = 55
-4 + 75 = 71
-3 + 100 = 97
Step-3: Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -3 and 100
10x^2 - 3x + 100x - 30
Step-4: Add up the first 2 terms, pulling out like factors:
x • (10x-3)
Add up the last 2 terms, pulling out common factors:
10 • (10x-3)
Step-5: Add up the four terms of step 4:
(x+10) • (10x-3)
Which is the desired factorization
Thus your answer is
Answer:
Option B.
Step-by-step explanation:
Let
b------> the number of buses
we know that
-----> inequality that represent the situation
Solve for b