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Oksanka [162]
3 years ago
9

Plsss! help me will mark brainlest!

Mathematics
2 answers:
jek_recluse [69]3 years ago
8 0

Answer:

Two polygons are similar with factor k.

=> The ratio between the area of two polygons: r = k^2

=> Here, the area of polygon B is larger than polygon A: 5^2 = 25 times

Hope this helps!

:)

GarryVolchara [31]3 years ago
7 0

Answer:

Area of polygon B is larger than the area of Polygon A by 25 times.

Step-by-step explanation:

The simplest way of finding this would be to draw a rough image of polygon A and B.

Let's say you drew Polygon A and B as rectangles. Since the scale factor is 5, the dimensions of the sides of polygon B would be 5 times as that of the dimensions of polygon A.

if you label the dimensions of Polygon A as length 2cm and width as 1cm, its area would be 2 square centimetres.

now keeping the scale factor of 5 in mind, multiply the length and the width by 5. you will get the length of polygon B as 10cm and width as 5cm. the area of polygon B becomes 50 square centimetres.

once you have done this, you will realise that the area of polygon B is 25 times the area of polygon A.

Hope this helped:)

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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}

3(31A+9B)=9 X 227000

31A+9B=681000

0.22A+0.05B=4500 (Multiply by 9)

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

8 0
3 years ago
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