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3 years ago

Help please,I dont know how do it!

Mathematics

1 answer:

nadezda [96]3 years ago

3 0

Use a formula given to you from there u will understand I am in 8th grade text me at 4042876133 for further help

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Simplify: {(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

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

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

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

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

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

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

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

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

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

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

$\sf 2^{nd} \: Law =$

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

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

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

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

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

3 0

2 years ago

Read 2 more answers

Solve 3x+ 4x = 2. (2 points) O 2t10 O-2+2v10 3 O-2±10 3 O -41v10 3

Norma-Jean [14]

3x+4x=2
You add 3x and 4x which gets you 7x

7x=2
You divide both sides of the equation by 7

x=2/7
This will be your result

3 0

3 years ago

You go for a run on monday you run 2.41Km on tuesday you ran the same distance how far did you ran altogether

FrozenT [24]

Answer: 4.82 km

Step-by-step explanation:

You went for a run on Monday.

You then went on a run on Tuesday as well. On this day you ran 2.41 km and it is said that you ran the same distance on both days which means you ran for 2.41 km on Monday as well.

The total distance you ran is therefore:

= 2.41 + 2.41

= 4.82 km

4 0

3 years ago

Cual es el 20% de 75

ValentinkaMS [17]

20% de 75 es 50...Es el iniciador

4 0

2 years ago

Read 2 more answers

subset to real numbers, and as you can see are not at all connected to the rest of numbers (and integers).

Natali5045456 [20]

Irrational numbers are the subset of real numbers that are not at all connected to the rest of numbers.

The subsets of real numbers are natural numbers, whole numbers, integers, rational numbers and irrational numbers.

Natural numbers are subset of whole numbers, which are subset of integers, which are subset of rational numbers. Hence, all of them are interconnected. The set of irrational numbers is the only subset of real numbers which is not associated with the rest.

For example:-

1 is a natural number, whole number, integer, rational number but not irrational number. On the other hand, $\sqrt{2}$ is an irrational number but none of the rest.

To learn more about real numbers, here:-

brainly.com/question/551408

#SPJ4

6 0

1 year ago