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GREYUIT [131]
2 years ago
13

Over 2 years, how much more does $2000 in a savings account with an APR of 4.2% compounded semiannually earn in interest than th

e same amount in a savings account with an APR of 3.6% compounded monthly? A. $24.29 B. $12.15 C. $6.07 D. $48.58
Mathematics
2 answers:
Reptile [31]2 years ago
7 0
Below is the solution:

 <span>A = 2000[1 + (.042/2)]^(2*2) 
A = 2173.366476962 
A = 2173.37 
I = 2173.37 - 2000 
I = 173.37 

A = 2000[1 + (.036/12)]^(12*2) 
A = 2149.0790382667085294501721261072 
A = 2149.08 
I = 2149.08 - 2000 
I = 149.08 

173.37 - 149.08 = 24.29</span>
-Dominant- [34]2 years ago
4 0

Answer:

$24.29 is correct

Step-by-step explanation:

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In a right triangle, angle A measures 20°. The side opposite angle A is 10 centimeters long. Approximately how long is the hypot
lubasha [3.4K]
Given that the angle measure 20 and the side opposite to that angle measures 10 cm, suppose this is the height of the triangle, the hypotenuse
Such that
sin theta=opposite/hypotenuse
opposite=a=10 cm
sin 20=10/h
multiplying both sides by h we get
hsin20=10
hence;
h=10/sin20
h=29.24 cm
h=29.2 cm 

3 0
2 years ago
Read 2 more answers
An employee at a homemade wooden toy store earned $860 over the past week. The employee needs to pay 14% for Federal Income Tax
kari74 [83]

Answer:

$713.8

Step-by-step explanation:

14%of 860 = 120.4

3% of 860 = 25.8

so, subtract the sum of the above 2 values from 860.

860 - (120.4+25.8) = 860-146.2 = 713.8

Hope this helps

3 0
2 years ago
Hospital sends an invoice to a patient. The patient schedules a payment plan in which she makes an initial payment of $1,451 the
Korvikt [17]
First month she payed 1451 dollars

Second month she payed 1/3 of that, which is:
1451/3 = 483.66

Third month she payed 1/3 of 483.66 because as text says, every next month she pays 1/3 of previous month payed amount.

483.66/3 = 161.22

Forth month she payed

161.22/3 = 53.74

In total she payed:

$2149.62 - B.

hope this helps :)
3 0
2 years ago
Read 2 more answers
Which statement identifies how to show that j(x) = 11.6ex and k(x) = In (StartFraction x Over 11.6 EndFraction) are inverse func
Vadim26 [7]

Answer:

<h2>It must be shown that both j(k(x)) and k(j(x)) equal x</h2>

Step-by-step explanation:

Given the function  j(x) = 11.6e^x and k(x) = ln \dfrac{x}{11.6}, to show that both equality functions are true, all we need to show is that both  j(k(x)) and k(j(x)) equal x,

For j(k(x));

j(k(x)) = j[(ln x/11.6)]

j[(ln (x/11.6)] = 11.6e^{ln (x/11.6)}

j[(ln x/11.6)] = 11.6(x/11.6) (exponential function will cancel out the natural logarithm)

j[(ln x/11.6)] = 11.6 * x/11.6

j[(ln x/11.6)] = x

Hence j[k(x)] = x

Similarly for k[j(x)];

k[j(x)] = k[11.6e^x]

k[11.6e^x] = ln (11.6e^x/11.6)

k[11.6e^x]  = ln(e^x)

exponential function will cancel out the natural logarithm leaving x

k[11.6e^x]  = x

Hence k[j(x)] = x

From the calculations above, it can be seen that j[k(x)] =  k[j(x)]  = x, this shows that the functions j(x) = 11.6e^x and k(x) = ln \dfrac{x}{11.6} are inverse functions.

4 0
2 years ago
Quadrilateral ABCD is translated 7 units down and 2 units to the right.
Inessa05 [86]
It will remain the same. 
6 0
2 years ago
Read 2 more answers
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