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3 years ago

What is five times three hundred

Mathematics

1 answer:

spayn [35]3 years ago

4 0

The answer is 1500 I hope this help!

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The probability of getting a number between 1 and 100 which is divisible by 3 is

Arlecino [84]

There are 33 numbers between 1 and 100 divisible by 3, therefore the probability is 33% or 1/3

5 0

3 years ago

A pool table is 8 feet long and 4 feet wide. How far is it from one corner pocket to the diagonally opposite corner pocket?

Mashutka [201]

Answer: 8.94427

Step-by-step explanation: using the Pythagorean theorem which is A squared plus B squared equals C squared you can find this

3 0

3 years ago

What is the first step in solving the equation x2 - 25 = 0?

Finger [1]

Answer:

first step - make x² the subject of the formula

x² = 25

second step :

find the square root of both sides

√x² = √25

x = 5

Step-by-step explanation:

5 0

2 years ago

Frank's been traveling a lot lately as part of his job. During the past 30 days, he's been out of town 18 days. What fraction re

siniylev [52]

Answer:

18/30

Step-by-step explanation:

He has been out of town 18 out of the 30 days. You can simplify this to 6/10.

3 0

2 years ago

Look at this cylinder:

Sedbober [7]

Height=h=8cm
Radius=r=4cm

We know

$\boxed{\sf \star TSA_{(Cylinder)}=2\pi r(h+r)}$

$\\ \sf\longmapsto TSA_{(Old\:Cylinder)}=2\times \dfrac{22}{7}\times 4(8+4)$

$\\ \sf\longmapsto TSA_{(Old\:Cylinder)}=\dfrac{176}{7}(12)$

$\\ \sf\longmapsto TSA_{(Old\:Cylinder)}=\dfrac{2112}{7}$

$\\ \sf\longmapsto TSA_{(Old\:Cylinder)}=301.7cm^2$

Now

New Radius=2(4)=8cm
New Height=2(8)=16cm

$\\ \sf\longmapsto TSA_{(New\:Cylinder)}=2\times \dfrac{22}{7}\times 8(16+8)$

$\\ \sf\longmapsto TSA_{(New\:Cylinder)}=\dfrac{352}{7}(24)$

$\\ \sf\longmapsto TSA_{(New\:Cylinder)}=\dfrac{8448}{7}$

$\\ \sf\longmapsto TSA_{(New\:Cylinder)}=1204.7cm^2$

$\\ \sf\longmapsto \dfrac{TSA_{(New\:Cylinder)}}{TSA_{(Old\:Cylinder)}}=\dfrac{1204.7}{301.7}$

$\\ \sf\longmapsto \dfrac{TSA_{(New\:Cylinder)}}{TSA_{(Old\:Cylinder)}}=\dfrac{4}{1}$

$\\ \sf\longmapsto\underline{\boxed{\bf{ {TSA_{(New\:Cylinder)}}:{TSA_{(Old\:Cylinder)}}=4:1}}}$

Hence our correct option is Option C

6 0

2 years ago