All categories

3 years ago

How do you solve for total intrest when you finance $600 for 24 months at 18%

Mathematics

2 answers:

boyakko [2]3 years ago

7 0

I = PRT (Principal, Rate, Time)
I = ?
P = 600
R = .18
T (years) = 2
600 x .18 x 2 is 216
the intrest is 216

yawa3891 [41]3 years ago

4 0

Multiple 18% by 600 and then multiply the number you get from that by 24 and add that number onto 600 and that’s your answer

You might be interested in

A/m=4g-h <br>need help with this problem

Olegator [25]

Answer:

a= m*(4g-h)

Step-by-step explanation:

hope it helps you

4 0

2 years ago

Read 2 more answers

What is 1 ft^3 to m^3

cluponka [151]

Answer:

1 ft3= 0,028 m3

Step-by-step explanation:

We know that a 1 ft= 0,305 m,

if we cubed the above equation we have the following

1 ft3= (0,305)^3 m3

1 ft3= 0,028 m3

5 0

3 years ago

Consider the region bounded by 4y=x^2 and 2y=x.

gayaneshka [121]

Answer:

a) ⅓ units²

b) 4/15 pi units³

c) 2/3 pi units³

Step-by-step explanation:

4y = x²

2y = x

4y = (2y)²

4y = 4y²

4y² - 4y = 0

y(y-1) = 0

y = 0, 1

x = 0, 2

Area

Integrate: x²/4 - x/2

From 0 to 2

(x³/12 - x²/4)

(8/12 - 4/4) - 0

= -⅓

Area = ⅓

Volume:

Squares and then integrate

Integrate: [x²/4]² - [x/2]²

Integrate: x⁴/16 - x²/4

x⁵/80 - x³/12

Limits 0 to 2

(2⁵/80 - 2³/12) - 0

-4/15

Volume = 4/15 pi

About the x-axis

x² = 4y

x² = 4y²

Integrate the difference

Integrate: 4y² - 4y

4y³/3 - 2y²

Limits 0 to 1

(4/3 - 2) - 0

-2/3

Volume = ⅔ pi

7 0

3 years ago

A calculator is required to obtain the final answer on this question. A solid metal sphere at room temperature 20oC is dropped i

yKpoI14uk [10]

Answer:

T = 47.1875°C

Step-by-step explanation:

Given:

Surrounding temp, Ts = 100°C

Initial Temperature ,T0= 20°C

Increase in temperature = 15°C

Final temperature, T = 20 + 15 = 35°C

Time, t = 9 seconds

Let's take Newton's law of cooling:

$T - Ts = (T_0 - Ts)e^k^t$

We'll solve for k

$35 - 100 = (20 - 100)e^9^k$

$-65 = (-80)e^9^k$

Divide both sides by -5

$13 = (16)e^9^k$

$9k = ln (\frac{13}{16})$

$k = \frac{1}{9} ln (\frac{13}{16})$

$k = -0.02307104053$

Let's now find the temperature of the ball after 18 seconds in boiling water.

Use the Newton's equation again:

$T - Ts = (T_0 - Ts)e^k^t$

$T - 100 = (20 - 100)e^-0.0^2^3^0^7^1^0^4^0^5^3^*^1^8$

$T - 100 = (-80)e^-0.0^2^3^0^7^1^0^4^0^5^3^*^1^8$

$T - 100= -52.8125$

$T = 100 - 52.8125$

T = 47.1875°C

Temperature of the ball after 18 seconds in boiling water is 47.1875°C

5 0

3 years ago

For 20 points, 5 stars, and a thanks, can some one answer these 5 questions about percent increase and decrease?

Galina-37 [17]

Answer:

380; 5% increase is 399

39; 33% decrease 26.13

48; 15% increase 55.2

3,000; 325% increase 12,750

160; 20% decrease 128

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers