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3 years ago

A down payment of 15% of the $21,200 purchase price of a new car is made. Find the amount financed.

1 answer:

6 0

Hello.

The answer to 1 is $3,180
All you have to do is divide 21,200 by 15%

The answer to 2 that is also true. You would have to find the unit rate to be able to find how much miles to operate.

Have a nice day

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A painting is 15 in tall and 20 in wide. If it is reducted to a width of 4 in, then how tall will it be?

inn [45]

20/4=5
15/5=3
3in
(However, it is not specified whether it is also compressed in it’s height as well as it’s length - just stick with the 3 inch height)

7 0

2 years ago

Eight more than one - fourth of d

alexandr1967 [171]

Answer:

(1/4*d) + 8

Sorry but there's no explanation cause its quick

5 0

3 years ago

Bradley is purchasing a new car for 37000. If the car loses 20%of its value every year, how much will the car be worth in 7 year

spayn [35]

ANSWER

After 7 years the car will worth, 7,759.5

EXPLANATION

The initial value of the car is 37000.

The rate of depreciation is 20%

In 7 years time, we can calculate the value of the car using the formula,

$A=P(1-r\%)^n$

We substitute the values into the formula to obtain,

$A=37000(1-0.2)^7$

$A=37000(0.88)^7$

$A=7759.5$ to the nearest tenth

6 0

2 years ago

Nita uses 1/3 of a carton of 12 eggs how many eggs does she use

Valentin [98]

What's 1/3 of 12?

Mulitply 1/3 times 12 = 4

Nita uses 4 eggs.

5 0

2 years ago

What is the fourth term in the binomial expansion (a+b)^6)

Dafna11 [192]

Answer:

$20a^3b^3$

Step-by-step explanation:

Binomial Series

$(a+b)^n=a^n+\dfrac{n!}{1!(n-1)!}a^{n-1}b+\dfrac{n!}{2!(n-2)!}a^{n-2}b^2+...+\dfrac{n!}{r!(n-r)!}a^{n-r}b^r+...+b^n$

Factorial is denoted by an exclamation mark "!" placed after the number. It means to multiply all whole numbers from the given number down to 1.

Example: 4! = 4 × 3 × 2 × 1

Therefore, the fourth term in the binomial expansion (a + b)⁶ is:

$\implies \dfrac{n!}{3!(n-3)!}a^{n-3}b^3$

$\implies \dfrac{6!}{3!(6-3)!}a^{6-3}b^3$

$\implies \dfrac{6!}{3!3!}a^{3}b^3$

$\implies \left(\dfrac{6 \times 5 \times 4 \times \diagup\!\!\!\!3 \times \diagup\!\!\!\!2 \times \diagup\!\!\!\!1}{3 \times 2 \times 1 \times \diagup\!\!\!\!3 \times \diagup\!\!\!\!2 \times \diagup\!\!\!\!1}\right)a^{3}b^3$

$\implies \left(\dfrac{120}{6}\right)a^{3}b^3$

$\implies 20a^3b^3$

7 0

1 year ago