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uranmaximum [27]

3 years ago

11

Sue has an annual interest income of $3390 from two investments. She has $10,000 more invested at 8% than she has at 6%. Find th

e amount invested at each rate.

1 answer:

olchik [2.2K]3 years ago

7 0

Answer:

$18500 at 6% and $28500 at 8%

Step-by-step explanation:

Let x represent the amount invested at 6%. Then x+10000 is the amount invested at 8%. The total income from the two investments is the sum of the products of the amount invested and the interest rate:

x·6% +(x+10000)·8% = 3390

0.14x + 800 = 3390 . . . . . . . . collect terms

0.14x = 2590 . . . . . . . . . . . . . .subtract 800

2590/0.14 = x = 18500 . . . . . divide by the coefficient of x

Sue has invested $18500 at 6% and $28500 at 8%.

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Consider the sequence defined recursively by do = 0, an = an-1 + 3n – 1. a) Write out the first 5 terms of this sequence,

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Answer:

The first 5 terms of the sequence is 2,7,15,26,40.

Step-by-step explanation:

Given : Consider the sequence defined recursively by $a_0=0$ $a_n=a_{n-1}+3n-1$

To find : Write out the first 5 terms of this sequence ?

Solution :

$a_n=a_{n-1}+3n-1$ and $a_0=0$

The first five terms in the sequence is at n=1,2,3,4,5

For n=1,

$a_1=a_{1-1}+3(1)-1$

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For n=2,

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$a_2=a_{1}+6-1$

$a_2=2+5$

$a_2=7$

For n=3,

$a_3=a_{3-1}+3(3)-1$

$a_3=a_{2}+9-1$

$a_3=7+8$

$a_3=15$

For n=4,

$a_4=a_{4-1}+3(4)-1$

$a_4=a_{3}+12-1$

$a_4=15+11$

$a_4=26$

For n=5,

$a_5=a_{5-1}+3(5)-1$

$a_5=a_{4}+15-1$

$a_5=26+14$

$a_5=40$

The first 5 terms of the sequence is 2,7,15,26,40.

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