Function can be defined as an expression, rule, or law that defines a relationship between one variable known as the independent variable and another called the dependent variable.

From the information given ,we have;

f(x)= 6x+7, with f ( r - 2)

From here, we have that x = r -2

Now, let's substitute x as r - 2 in the function given;

f(x) = 6x + 7

f(r - 2) = 6 ( r - 2) + 7

Expand the bracket

f(r - 2) = 6r - 12 + 7

collect like terms

f(r - 2) = 6r - 5

Thus, the value of f(r - 2) is 6r - 5

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5 0

2 years ago

What is the equation of a line that passes through (1,2) and has a slope of 7?

Paraphin [41]

ANSWER: y=7x-5 i put it in a graphing calculator

3 0

3 years ago

Which of these choices show a pair of equivalent expressions? Check all that apply.

AleksAgata [21]

The main rule we use in this exercise is

$( \sqrt{a} ) ^{m} =( a^{ \frac{1}{2}} )^{m} = a^{ \frac{1}{2}*m}=a^{ \frac{m}{2}}$

and the more general case:

$( \sqrt[n]{a} ) ^{m} =( a^{ \frac{1}{n}} )^{m} = a^{ \frac{1}{n}*m}=a^{ \frac{m}{n}}$
Using these rules, we make all the indices, or exponents, look like $a^{ \frac{m}{n}}$ , and compare them to each other.

A. $( \sqrt{8} )^{9} = ( 8^{ \frac{1}{2}} )^{9} = 8^{ \frac{1}{2}*9}= 8^{ \frac{9}{2}}$

B. $( \sqrt{4} ) ^{5} =( 4^{ \frac{1}{2}} )^{5} = 4^{ \frac{1}{2}*5}=4^{ \frac{5}{2}}$

C. $( \sqrt[3]{125} ) ^{7} =( 125^{ \frac{1}{3}} )^{7} = 125^{ \frac{1}{3}*7}= 125^{ \frac{7}{3}}$

D. $( \sqrt{12} ) ^{7} =( 12^{ \frac{1}{2}} )^{7} = 12^{ \frac{1}{2}*7}=12^{ \frac{7}{2}}$

4 0

4 years ago

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