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
Thus, the stockbroker invested $300 at 7%, $600 at 8%, and $900 at 9%.
After you re-arrange the equation, you'll find out that you got y=3x+6....okay, this equation is a linear equation, you can use y=mx+b {where m=3, and b=6}. then to find the co-ordinate of (x,y) which is the point, substitute 1,2,3,4.... for x in 'y=3x+6' to figure out the co-ordinate of y.
answer: (1,14), (2,16), (3,19), (4,22)
Answer:
tan (A-B) = ± 4/3
Step-by-step explanation:
COS (A-B) = 3/5
COS² (A-B) = (3/5)² = 9/25 = 1 - sin² (A-B)
sin² (A-B) = 1 - 9/25 = 16/25
sin (A-B) = ± 4/5
tan (A-B) = sin (A-B) / cos (A-B) = (± 4/5) / (3/5) = ± 4/3
The answer is: [B]: " 9 * 10⁻³ " .
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Answer:
-6x-19
Step-by-step explanation:
