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DIA [1.3K]
3 years ago
14

How many times can 5 go into 1620

Mathematics
1 answer:
kaheart [24]3 years ago
5 0
1620/ 5 is =324
To check just multiply
5 * 324
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skate land charges a $50 fee for a birthday party rental and $4 for each person. Joann has no more than $100 to budget for her p
otez555 [7]

Answer:

11 then Joann would have $2 left over.

7 0
3 years ago
A plane flew a total of 2220 miles its average speed was 555 mph how many hours did the plane fly
o-na [289]
2220 miles divided by 554mph would give you 4 hours
5 0
3 years ago
240 fluid ounces = ____ cups
katen-ka-za [31]

Answer:

30

Step-by-step explanation:

for every 1 fluid once it is 0.125 cup. So keep adding up and you will get to 240 and the amount will be 30.

8 0
3 years ago
A cylinder and cone have the same height and radius. The height of each is 5 cm, and the radius is 2 cm. Calculate the volume of
never [62]

Answer:

The volume of a cylinder is 62.8 cm and the volume of a cone is 20.9 cm³ .

Step-by-step explanation:

Formula

Volume\ of\ a\ cylinder = \pi r^{2} h

Volume\ of\ a\ cone = \pi r^{2} \frac{h}{3}

Where r is the radius and h is the height .

As given

A cylinder and cone have the same height and radius. The height of each is 5 cm, and the radius is 2 cm.

\pi = 3.14

Thus

Volume\ of\ a\ cylinder =3.14\times 5\times 2\times 2

                                             = 62.8 cm³

Thus the volume of a cylinder is 62.8 cm³ .

Volume\ of\ a\ cone = 3.14\times 2\times 2\times frac{5}{3}

Volume\ of\ a\ cone =\frac{5\times 3.14\times 2\times 2}{3}

Volume\ of\ a\ cone =\frac{62.8}{3}

                                        = 20.9 (Approx) cm³

Thus the volume of a cone is 20.9 cm³ .

Therefore the volume of a cylinder is 62.8 cm and the volume of a cone is 20.9 cm³ .

8 0
3 years ago
Please help with algebra problem
kumpel [21]
Number of weekend minutes used: x
Number of weekday minutes used: y

This month Nick was billed for 643 minutes:
(1) x+y=643

The charge for these minutes was $35.44
Telephone company charges $0.04 per minute for weekend calls (x)
and $0.08 per minute for calls made on weekdays (y)
(2) 0.04x+0.08y=35.44

We have a system of 2 equations and 2 unkowns:
(1) x+y=643
(2) 0.04x+0.08y=35.44

Using the method of substitution
Isolating x from the first equation:
(1) x+y-y=643-y
(3) x=643-y

Replacing x by 643-y in the second equation
(2) 0.04x+0.08y=35.44
0.04(643-y)+0.08y=35.44
25.72-0.04y+0.08y=35.44
0.04y+25.72=35.44

Solving for y:
0.04y+25.72-25.72=35.44-25.72
0.04y=9.72

Dividing both sides of the equation by 0.04:
0.04y/0.04=9.72/0.04
y=243

Replacing y by 243 in the equation (3)
(3) x=643-y
x=643-243
x=400

Answers:
The number of weekends minutes used was 400
The number of weekdays minutes used was 243
6 0
3 years ago
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