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devlian [24]
2 years ago
11

(a+b)^2–(b+c)^2 as a product of two polynomials

Mathematics
1 answer:
vovikov84 [41]2 years ago
8 0

Hey there!!

In order to solve this problem, we will need to use the identity a² - b²

∴ a² - b² = ( a + b ) ( a - b )

According to the question,

a = a + b

b = b + c

( a + b )² - ( b + c )²

... ( a + b + b + c ) ( a + b - b - c )

... ( a + 2b + c ) ( a - c )

Now, we will need to solve this using distributive property.

Distribute a with a , 2b , c and distribute -c with a , 2b and c and then combine all the like terms.

... ( a² + 2ab + ac ) + ( -ac - 2bc - c² )

... ( a² + 2ab + ac - ac - 2bc - c² )

... ( a² + 2ab - bc - c² )

Hope my answer helps!!

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A=1/10, b=1/12, c=1/15  (these are the rates for each worker...)

1.)

2(1/10)+4(1/12)+t/15=1

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So it would take c 7 hours to finish the job.

2.)

1/40-1/60

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1. Taylor needs to purchase a car. The car Taylor plans to purchase costs $10,000. Taylor has saved $2,000 to use as a down paym
Alexxandr [17]

Answer:

a. The amount Taylor will need to finance is $8,000

b. The amount Taylor pays as interest in one year is $400

Step-by-step explanation:

The given parameters of the financing for the car are;

The cost of the car Taylor plans to purchase, C = $10,000

The amount Taylor has saved to be used as down payment, S = $2,000

The interest rate of the credit Taylor is offered = 5%

The duration given for repayment of the loan = 5 years

a. To purchase the car, the amount Taylor will need to finance, 'P', is given as follows;

P = C - S

∴ P = $10,000 - $2,000 = $8,000

b. The amount of interest on the loan in one year, 'I', is given by the following formula;

I = \dfrac{P \times R \times T}{100}

Where;

I = The interest payed

P = The principal amount taken as loan = $8,000

R = The interest rate = 5% APR

T = The time period the interest is applied = 1 year

Plugging in the values, we get;

I = \dfrac{\$ \, 8,000 \times 5 \times 1}{100} = \$ \, 400

The interest Taylor will pay on the loan in one year, I = $4,00

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B. indefinitely many

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