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sweet [91]
3 years ago
15

Ernie Rolph borrowed ​$3200 at 10​% annual simple interest. If exactly 1 year later he was able to repay the loan without​ penal

ty, how much interest would he​ owe?
Mathematics
1 answer:
Setler [38]3 years ago
6 0

Answer:

$320

Step-by-step explanation:

From simple interest formular

A=P(1+int) where A is amount payable, P is principal amount borrowed and int is interest gained

int=Rt where R is rate of interest in decimal form, t is duration

The rate of 10% converted to decimal is 10/100=0.1

Substituting P for $3200, R for 0.1 and t for 1 year

A=$3200(1+(0.1*1))=$3200*(1+0.1)=3200*1.1=$3520

To find the interest, it's A-P hence $3520-$3200=$320

Therefore, interest owed is $320

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4 m<br> 3 m<br> What is the length of the hypotenuse?
wlad13 [49]

Answer:

H^2 = 4^2 + 3^2

H^2 = 16 + 9

H^2 = 25

H = √25

H = 5

5 0
3 years ago
5(4x+9)=-50+15 solve the equation
Alenkinab [10]

Answer:

x = -4

Step-by-step explanation:

5(4x+9)=-50+15

20x+45=-50+15

20x+45=-35

      -45  -45

20x=-80

/20   /20

x=-4

7 0
3 years ago
ONLY SERIOUS ANSWERS!!!! 25 POINTS
AveGali [126]

Answer:

FD≈25.94.. rounded = 26

Step-by-step explanation:

FD²=12²+(4x+11)²

FD²=144+16x²+88x+121

FD²=265+16x²+88x

also

FD²=12²+(13x-16)²

FD²=144+169x²-416x+256

FD²=400+169x²-416x

thus

265+16x²+88x = 400+169x²-416x

16x²-169x²+88x+416x+265-400 = 0

-153x²+504x-135 = 0

we will solve this quadratic equation by suing the quadratic formula to find x

x=(-504±sqrt(504²-4(-153)(-135)))/2(-153)

x=(-504±\sqrt{504^{2}-4(-153)(-135) })/2(-153)

x=(-504±\sqrt{254016-82620 })/-306

x=(-504±\sqrt{171396})/-306

x=(-504±414)/-306

x=(-504+414)/-306 and x=(-504-414)/-306

x=-90/-306 and x=-918/-306

x= 5/17 , 3

substituting x by the roots we found

check for 5/17:

4x+11 = 4×(5/17)+11 = (20/17)+11 = (20+187)/17 = 207/17 ≈ 12.17..

13x-16 = 13×(5/17)-16 = (65/17)-16 = (65-272)/17 = -207/17 ≈ -12.17..

check for 3:

4x+11 = 4×3+11 = 12+11 = 23

13x-16 = 13×3-16 = 23

thus 3 is the right root

therfore

ED=23 and CD=23

FD²=FE²+ED² or FD²=FC²+CD²

FD²=12²+23²

FD²=144+529

FD²=673

FD=√673

FD≈25.94.. rounded = 26

5 0
3 years ago
Read 2 more answers
13[6^2\(5^2-4^2)+9] simplified form of the expression
dmitriy555 [2]

13[6^2/(5^2-4^2)+9

(25-16)

13[36/9+9]

4+9

13*13

169

8 0
3 years ago
The minimum of a parabola is located at (-1,-3). the point (0,1) is also on the graph. which equation can be solved to determine
mario62 [17]

Answer:

The equation that can be used to solve for a is 1 = a(0=1)²-3.

Explanation:

In this case, I would model the parabola in vertex form.  Vertex form is y = a(x-h)²+k, where (h,k) is the vertex of the parabola.  Using the information from the question, we can plug in values to get 1 = a(0+1)²-3.  (This could be simplified as 1 = a(1)²-3 ⇒ 1 = a-3 ⇒ a = 4, but we are only interested in finding the equation that can solve for a.)

8 0
4 years ago
Read 2 more answers
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