1. 2x2 – x – 6
2. 12x2 – x –6
3. 4p2 – 18p + 8
Answer:
Kindly rewrite the question clearly.
Answer:
- $5000 at 10%, $10000 at 12% and 10000 at 16%
Step-by-step explanation:
- <em>One part of $ 25,000 is invested at 10% interest, another part at 12%, and the rest at 16%. The total annual income from the three investments is $ 3,200. Also, the income from the investment at 16% is equal to the income from the other two investments combined. How much was invested at each interest rate?</em>
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Let the parts be x, y and z
<u>As per given we get below system of equations:</u>
- x + y + z = 25000
- 0.1x + 0.12y + 0.16z = 3200
- 0.1x + 0.2y = 0.16z
<u>Substitute 0.1x + 0.2y in the second equation:</u>
- 0.16z + 0.16z = 3200
- 0.32z = 3200
- z = 3200/0.32
- z = 10000
<u>Now we have:</u>
- x + y + 10000 = 25000 ⇒ x + y = 15000
and
- 0.1x + 0.12y + 0.16*10000 = 3200 ⇒ 0.1x + 0.12y = 1600
<u>Multiply the second equation and then subtract the first one:</u>
- 10(0.1x + 0.12y) = 10(1600) ⇒ x + 1.2y = 16000
- x + 1.2y - (x + y) = 16000 - 15000
- 0.2y = 1000
- y = 10000
Then
<u>So the parts are:</u>
- $5000 at 10%, $10000 at 12% and 10000 at 16%
Answer: What do I solve for?
Step-by-step explanation:
Do I need to solve for the slope, equation, x-int, etc?
Ava's sister paid a 4.5% annual interest rate
Her sister paid $54 more than she borrowed, divide that by 3 (years till she paid)
54/3 = 18
That's $18 for every year she'd had the $400 loan out, divide it by 4 to get the percentage
18/4 = 4.5/100 = 4.5%
