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Minchanka [31]
2 years ago
6

If someone solved this can you help?

Mathematics
1 answer:
natka813 [3]2 years ago
5 0

Answer:

I/d/k but I will say 9/10

Step-by-step explanation:

Wow

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You owe $10,000 on a car loan. The bank charges a compound interest rate of 9.5% per year. If you don't make any payments on the
kipiarov [429]
Ya

compound interest
A=P(1+r)^t
A=amount final
P=principal
r=rate in decimal
t=time in yers

so

interest is 9.5% or 0.095
principal is 10000
time is 3 years
A=10000(1+0.095)^3
A=10000(1.095)^3
A=13129.323
rounded
he owes $13129.32
7 0
3 years ago
Which of the numbers below could be terms in the sequence an = 4n - 7? Check all that apply.
cluponka [151]
There are several possible terms which can be included given the sequence formula an = 4n - 7. If we list all the possibilities, here are as follows:

n      an
0      -7
1      -3
2      1
3      5
.       .
.       .
.       .
n      an

Substitute any number n to an and you can get all the terms in the sequence.
6 0
3 years ago
Read 2 more answers
What is 7+20 answer quick
Zinaida [17]

Answer: 27

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
1. Identify the RATE OF CHANGE and INITIAL VALUE from
Jet001 [13]

Here, Rate of change means Slope and initial value means y-intercept. The equations given above are written in slope-intercept intercept form i.e.

\bf \: y = mx + b

Where,

  • m is slope
  • b is the y-intercept

\tt \: y = 3x + 6

  • Rate of change = 3
  • Initial value = 6

\tt \: y =  - 7x + 5

  • Rate of change = -7
  • Initial value = 5

\tt \: y = 9x + 4

  • Rate of change = 9
  • Initial value = 4
6 0
2 years ago
Consider two markets: the market for cat food and the market for dog food. The initial equilibrium for both markets is the same,
umka21 [38]

Answer:

elasticity supply of dog food = 2.61

elasticity supply of cat food = 1.71

Step-by-step explanation:

The midpoint formula for elasticity is:

Elasticity = \frac{(Q2-Q1)/[(Q2+Q1)/2]}{(P2-P1)/[(P2+P1)/2]}

Point 1: Q = 39.0 and P = 5.50

Point 2: Q = 101.0 and P = 7.75

Elasticity\ supply\ of\ dog\ food = \frac{(101.0-39.0)/[101.0+39.0)/2]}{(7.75-5.50)/[(7.75+5.50)/2]}=2.61

Doing the same for the cat food:

Elasticity\ supply\ of\ cat\ food = \frac{(71.0-39.0)/[71.0+39.0)/2]}{(7.75-5.50)/[(7.75+5.50)/2]}=1.71

4 0
3 years ago
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