Given:
The rate of interest on three accounts are 7%, 8%, 9%.
She has twice as much money invested at 8% as she does in 7%.
She has three times as much at 9% as she has at 7%.
Total interest for the year is $150.
To find:
Amount invested on each rate.
Solution:
Let x be the amount invested at 7%. Then,
The amount invested at 8% = 2x
The amount invested at 9% = 3x
Total interest for the year is $150.

Multiply both sides by 100.


Divide both sides by 50.


The amount invested at 7% is
.
The amount invested at 8% is

The amount invested at 9% is

Therefore, the stockbroker invested $300 at 7%, $600 at 8%, and $900 at 9%.
Answer:
steps below
Step-by-step explanation:
f(x) = x⁷ – 6x⁶ + 8x⁵
f'(x) = 7x⁶ - 36x⁵ + 40x⁴
= x⁴ (7x² - 36x + 40) ... (1)
Critical point (x,y): x = 0 , x = (36 ± √(-36)²-4*7*40) / (2*7) = (18±2√11) / 7
x = 0 or x = 3.5 or x = 1.6
plug into (1): y = 0 , y = -394.3 , y = 10.1
extreme : <u>maximum (1.6 , 10.1)</u> <u>minimum (3.5 , -394.3)</u> ... turning points
x⁷ – 6x⁶ + 8x⁵ = x⁵ (x² - 6x + 8) = x⁵ (x-2) (x-4) =0
<u>real zeros: x = 0, x=2, x=4</u>
Answer:
2x^5/3
Step-by-step explanation:
Answer:
18r
Step-by-step explanation:
since they both have r as their variable, they're like terms, meaning you can combine them
so 13+5 = 18