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3 years ago

Find the slope of the line that passes through the following points. Write "undefined" if the slope is undefined.

Mathematics

1 answer:

Katyanochek1 [597]3 years ago

5 0

The slope of the line that passes through the following points is undefined

Solution:

Given that, We have to find the slope of line passing through two points

Given points are (-1, 5) and (-1, 6)

The slope of line is given by formula:

$m = \frac{y_2-y_1}{x_2-x_1}$

From given,

$(x_1, y_1) = (-1, 5)\\\\(x_2, y_2) = (-1, 6)$

Substituting the values we get,

$m = \frac{6-5}{-1-(-1)}\\\\m = \frac{1}{-1+1}\\\\m = \frac{1}{0}$

Thus the slope is undefined

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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

3 years ago

Read 2 more answers

If n is an integer and can be written in expanded form....

aksik [14]

Log(10^-3) = log(0.001) = -3, and there are 3 zeros inclusive of the 0 before the decimal.

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3 years ago

What does 3 over 8 simplify to

nirvana33 [79]

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3 years ago

The ratio of small dogs to big dogs at the park is 1:3. Which does not represent this ratio?

Karolina [17]

Answer:

C.2:4

Step-by-step explanation:

1*2=2

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2:6=2:4

not true so it is C.

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