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Maksim231197 [3]
3 years ago
13

What shape would have a cross section that could match that of a sphere?

Mathematics
1 answer:
lapo4ka [179]3 years ago
3 0
Cross sections of spheres are cicles. Therefore, if cut parallel to the ends, CYLINDER also has cross sections that match that of a circle.
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If £2000 is placed into a bank account that pays 3% compound interest per year, how much will be in the account after 2years?
AVprozaik [17]
3% = 3/100 = 0,03

When it's compount interest it's 1,03 because you gain 3% more from what you already have.

All you need to do
1,03•2000=2060$ for the first year

And then for the second year, you take the actual amount after one year of 3%

1,03•2060=2121,8

After two years, there will be 2121,8$
5 0
3 years ago
Read 2 more answers
Simplify: <br>{(12)^1 + (13)^-1}/[(1/5)^-2 × {(1/5)^-1 + (1/8)^-1}^-1]​
CaHeK987 [17]

Step-by-step explanation:

\underline{\underline{\sf{➤\:\:Solution}}}

\sf \dashrightarrow \:  \dfrac{ \left(\left(12 \right)^{ - 1}  + \left(13 \right)^{ - 1}  \right) }{\left( \dfrac{1}{5}\right) ^{ - 2}  \times\left( \left( \dfrac{1}{5}  \right) ^{ - 1}  +\left( \dfrac{1}{8}  \right) ^{ - 1}  \right) ^{ - 1}}

\sf \dashrightarrow \:  \dfrac{ \left(\dfrac{1}{12}  + \dfrac{1}{13} \right) }{\left( \dfrac{5}{1}\right) ^{ 2}  \times\left( \dfrac{5}{1}  + \dfrac{8}{1}   \right) ^{ - 1}}

  • LCM of 12 and 13 is 156

\sf \dashrightarrow \:  \dfrac{ \left(\dfrac{1 \times 13 = 13}{12 \times 13 = 156}  + \dfrac{1 \times 12 = 12}{13 \times 12 = 156} \right) }{\ \dfrac{25}{1} \times\left( \dfrac{5 + 8}{1}    \right) ^{ - 1}}

\sf \dashrightarrow \:  \dfrac{ \left(\dfrac{13}{156}  + \dfrac{12}{156} \right) }{\ \dfrac{25}{1} \times\left( \dfrac{13}{1}    \right) ^{ - 1}}

\sf \dashrightarrow \:  \dfrac{ \left(\dfrac{13 + 12}{156}  \right) }{\ \dfrac{25}{1} \times\dfrac{1}{13} }

\sf \dashrightarrow \:    \dfrac{25}{156} \div    \dfrac{25}{13}

\sf \dashrightarrow \:    \dfrac{ \cancel{25}}{156}  \times    \dfrac{13}{ \cancel{25} }

\sf \dashrightarrow \:     \dfrac{13}{156}

\sf \dashrightarrow \:     \dfrac{1}{12}

\sf \dashrightarrow \:    Answer =   \underline{\boxed{ \sf{ \dfrac{1}{12} }}}

━━━━━━━━━━━━━━━━━━━━━━━━

\underline{\underline{\sf{★\:\:Laws\:of\: Exponents :}}}

\sf \: 1^{st} \: Law = \bigg( \dfrac{m}{n} \bigg)^{a} \times \bigg( \dfrac{m}{n} \bigg)^{b} = \bigg( \dfrac{m}{n} \bigg)^{a + b}

\sf 2^{nd} \: Law =

\sf Case : (i) \: if \: a > b \: then, \bigg( \dfrac{m}{n}\bigg) ^{a} \div \bigg( \dfrac{m}{n}\bigg)^{b} = \bigg( \dfrac{m}{n}\bigg)^{a - b}

\sf Case : (ii) \: if \: a < b \: then, \bigg( \dfrac{m}{n}\bigg) ^{a} \div \bigg( \dfrac{m}{n}\bigg)^{b} = \dfrac{1}{\bigg( \dfrac{m}{n}\bigg)^{b - a}}

\sf \: 3^{rd} \: Law = \bigg\{ \bigg( \dfrac{m}{n} \bigg)^{a} \bigg\}^{b} = \bigg( \dfrac{m}{n} \bigg)^{a \times b} =\bigg( \dfrac{m}{n} \bigg)^{ab}

\sf \: 4^{th} \: Law = \bigg( \dfrac{m}{n} \bigg)^{ - 1} = \bigg( \dfrac{n}{m} \bigg) =\dfrac{n}{m}

\sf \: 5^{th} \: Law = \bigg( \dfrac{m}{n} \bigg)^{0} = 1

3 0
2 years ago
Read 2 more answers
I have 30 ones, 2 thousands, 4 hundred thousands,60 tens and 100 hundreds. What number am I?
Pepsi [2]
Answer
The number is 412630
6 0
3 years ago
Need help ASAP !!!!<br> Solve for y
GarryVolchara [31]

Answer:

y=8.94

Step-by-step explanation:

use the formula: a^2+b^2=c^2

64+y^2=144

y=\sqrt{80}

or y=8.94

8 0
2 years ago
-9b-6=-3b +48<br> Step by step
Travka [436]

Answer:

b = -9

Step-by-step explanation:

1. Move the variables to the left side. Be sure to change the terms (add/subtract)

-9b-6 = -3b+48

-9b+3b-6 = 48

2. Combine like terms.

-9b+3b-6 = 48

-6b = 48 + 6

3. Divide both sides by -6.

-6b = 54

b = -9

7 0
3 years ago
Read 2 more answers
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