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Hatshy [7]
3 years ago
9

Please help, the question is in the picture

Mathematics
2 answers:
natita [175]3 years ago
5 0

Answer:

(i) 18a

(ii) 14a^2

Step-by-step explanation:

i: 7a+7a+2a+2a

      14a+4a

         18a

ii: 7a*2a

     14a2

I hope this helps!

jekas [21]3 years ago
3 0

Answer:

per = 18a         7a+2a+7a+2a

area = 14 a^{2}      7a*2a

Step-by-step explanation:

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the sherman family mad a down payment of $18,375 for a new house they mortaged the remaining cost of $129,625 the interest paid
Allisa [31]

Answer:

honestly this is a very poorly worded question...

it can be interpreted in several different ways....

the cost of the house (for them to purchase it) was

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Step-by-step explanation:

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3 years ago
50 POINTS FOR THE CORRECT ANSWER
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Answer:

Step-by-step explanation:

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Chelsea took out three loans for a total of ​$ 78 comma 000 78,000 to start an organic orchard. Her​ business-equipment loan was
lianna [129]

Answer:

Business-equipment loan, A=$15000

Small-business loan, B=$24000

home-equity loan, C= $39000

Step-by-step explanation:

Let the Business-equipment loan at an interest rate of​ 11%=A

Let the​ small-business loan was at an interest rate of 5​% =B

Let her​ home-equity loan was at an interest rate of 4.5​%.=C

Since her total Loan=$78000

A+B+C=78000....(I)

Simple Interest = (P X R X T)/100

Since the Time, T=1 year

Interest on A = 0.11A

Interest on B = 0.05B

Interest on C = 0.045C

The total simple interest due on the loans in one year was ​$4605.

0.11A+0.05B+0.045C=4605....(II)

The annual simple interest on the​ home-equity loan was ​$105 more than the interest on the​ business-equipment loan.

0.045C = 0.11A +105...(III)

We proceed to solve the simultaneous equations.

A+B+C=78000....(I)

0.11A+0.05B+0.045C=4605....(II)

0.045C = 0.11A +105...(III)

From (III), 0.045C = 0.11A +105

Substitute 0.045C = 0.11A +105 into (II).

0.11A+0.05B+0.11A +105=4605

0.22A+0.05B=4500

From (III),

C=\frac{22A}{9}+\frac{7000}{3}

Substitute into (I)

A+B+\frac{22A}{9}+\frac{7000}{3}=78000

\frac{31A}{9}+B=78000-\frac{7000}{3}

\frac{31A}{9}+B=\frac{227000}{3}

\frac{31A+9B}{9}=\frac{227000}{3}

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31A+9B=681000

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31A+9B=681000 (Multiply by 0.05)

1.98A+0.45B=40500

1.55A+0.45B=34050

Subtracting

0.43A=6450

A=$15000

From (III)

C=\frac{22A}{9}+\frac{7000}{3}

C=\frac{22X15000}{9}+\frac{7000}{3} =39000

C=$39000

from (I)

A+B+C=78000

15000+B+39000=78000

B=$24000

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