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miv72 [106K]
3 years ago
14

Brads statement shows a previous balance of $2592.35, a payment of $75, and new transactions totaling $432. His APR 18.6%. What

is his new balance?
(A) 2988.37
(B) 3064.53
(C) 3140.93
(D) 3246.43
Mathematics
1 answer:
weeeeeb [17]3 years ago
8 0

Answer:

$2988.37

Step-by-step explanation:

Subtract the payment from the previous balance.  Convert the APR to a decimal.  Multiply the unpaid balance by this decimal.  Divide the product by 12 to calculate the interest added for this month.  Add the new amount owed, the interest, and the new transaction to determine the new balance.

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vodomira [7]

Bro I think you should search algerba calcautor and then you'll get the answers

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3 years ago
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I am having a test tomorrow about this can someone help me explain it?​
Amiraneli [1.4K]

Answer:

the answer would be 217 [degrees]

Step-by-step explanation:

Since m<1 and m<2 are next to each other, they are adjacent. Adjacent angles  must share a common side which is m<W. This means they have to be added.   m<1+m<2= m<XYZ

145+72=217

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3 years ago
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The sector COB is cut from the circle with center O. The ratio of the area of the sector removed from the whole circle to the ar
mafiozo [28]

Answer:

Ratio = \frac{R^2 - r^2 }{ r^2}

Step-by-step explanation:

Given

See attachment for circles

Required

Ratio of the outer sector to inner sector

The area of a sector is:

Area = \frac{\theta}{360}\pi r^2

For the inner circle

r \to radius

The sector of the inner circle has the following area

A_1 = \frac{\theta}{360}\pi r^2

For the whole circle

R \to Radius

The sector of the outer sector has the following area

A_2 = \frac{\theta}{360}\pi (R^2 - r^2)

So, the ratio of the outer sector to the inner sector is:

Ratio = A_2 : A_1

Ratio = \frac{\theta}{360}\pi (R^2 - r^2) : \frac{\theta}{360}\pi r^2

Cancel out common factor

Ratio = R^2 - r^2 : r^2

Express as fraction

Ratio = \frac{R^2 - r^2 }{ r^2}

6 0
3 years ago
Can somebody prove this mathmatical induction?
Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

\sum \limits_{j=1}^k2^j=2(2^k-1)

3 step:

Check for n=k+1 whether the statement

\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)

is true.

Start with the left side:

\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)

According to the 2nd step,

\sum \limits_{j=1}^k2^j=2(2^k-1)

Substitute it into the \ast

\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)

So, you have proved the initial statement

4 0
3 years ago
Two girls had 45.00 to spend. one spent 6 dollars less than twice the other. solve
Varvara68 [4.7K]

Let, money spend by jane = x

Maggie spend = 2x-6

So, x+2x-6 = 45

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So, jane spend $17 & Maggie spend $28

3 0
3 years ago
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