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2 years ago

15

What would the circumference be of a circle if the radius is 3 units

1 answer:

MAVERICK [17]2 years ago

7 0

If we have a circle with a radius of 3 we need to find the diameter of the circle. Remember radius is half of the diameter. So, the diameter is 6 units.

Use the circumference formula: C=3.14D

C=3.14(6)

C=18.84 units

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Write a number with two decimals places, that is bigger than 4/ 1/5 but smaller than 4.25

Mkey [24]

Answer:

4.16 , 4.17 , 4.18 , ... , 4.24

8 0

2 years ago

You want to purchase a new car in 55

Tanzania [10]

Answer:

Ans. you should deposit each month to end up with $38,000 the amount of $533.33 every month for 5 years at a APR of 6.5%

Step-by-step explanation:

Hi, first we have to convert all the data to monthly basis, that is, 5 years (5*12=60 months) and the rate of 6.5% APR offered by the bank (Monthly rate = 0.065/12=0,005666667 or 0.5667% monthly)

With that in mind, we need to solve for "A" the following equation.

$FV=\frac{A((1+r)^{n}-1) }{r}$

Where:

FV = Future value of the car

r = rate of return offered by the bank

n = number of periods that you are going to make the monthly deposit

That is:

$38,000=\frac{A((1+0,005667)^{60}-1) }{0,005667}$

$38,000=\frac{A(0,403599896)}{0,005667}$

$38,000=A(71,22351103)$

$A= 533,53$

Best of luck.

5 0

2 years ago

A training field is formed by joining a rectangle and two semicircles, as shown below. The rectangle is 96m long and 74m wide.

Oduvanchick [21]

Answer:

$11402.66 m^{2}$

Step-by-step explanation:

The width of rectangle is the diameter of the semi-circle part

Area of one semicircle is given by $\frac {0.5\pi d^{2}}{4}$

Total area of semi circle will be $2\times\frac {0.5\pi d^{2}}{4}$

Substituting 74 m for d and $\pi$ as 3.14 we obtain

Total area semi-circle= $2\times\frac {0.5*3.14\times 74^{2}}{4}=4298.66 m^{2}$

Area of rectangle is given by the product of length and width

Rectangular area= $96 m*74 m=7104 m^{2}$

Total area of rectangular and semi-circles will be

$4298.66 m^{2}+7104 m^{2}=11402.66 m^{2}$

Therefore, area of training field is $11402.66 m^{2}$

6 0

2 years ago

I REALLY NEED HELP WITH MY ALEGBRA HOMEWORK!!!!!!!!

mamaluj [8]

Answer:

1. is 55 cans I dont know the fomula but ...

2. the 16th term is 521

3. Hope this help:D

Step-by-step explanation:

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2 years ago

In a lab, a substance was cooled by 24 C over a period of 6 hours at a constant rate. What was the change in temperature each ho

Karo-lina-s [1.5K]

Step-by-step explanation:

Decrease of 24°C in 6 hours

=> -24°C / 6 hours

=> -4°C / hour

6 0

2 years ago