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almond37 [142]
3 years ago
10

Write the equation of the line that passes through (2, 3) and is parallel to the line 12x – 5y = 2.

Mathematics
1 answer:
vladimir1956 [14]3 years ago
7 0
k:12x-5y=2\to-5y=-12x+2\ \ \ /:(-5)\to y=\frac{12}{5}y-\frac{2}{5}\\\\l:y=mx+b\\\\k\ ||\ l\iff m=\frac{12}{5}\\\\ l:y=\frac{12}{5}x+b;\ (2;\ 3)\\\\\frac{12}{5}\cdot2+b=3\\\\\frac{24}{5}+b=3\\\\b=\frac{15}{5}-\frac{24}{5}\\\\b=-\frac{9}{5}\\\\Answer:y=\frac{12}{5}x-\frac{9}{5}\to\frac{12}{5}x-y=\frac{9}{5}\ \ \ \ /\cdot5\to12x-5y=9
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What is the difference x/x^2-16-3/x-4
kenny6666 [7]

General Idea:

When simplifying a rational expression, we need to do the below steps:

(i) Factor the Denominator of each fraction

(ii) Identify the Least Common Denominator (It is the product of prime factors involved with its highest exponent)

(iii) Identify and rewrite the equivalent fraction with the desired LCD.

(iv) Once the denominator are same, Combine the numerator.

Applying the concept:

What is the difference x/x^2-16-3/x-4

I assume that you mean to type the expression \frac{x}{x^{2}-16}  -\frac{3}{x-4}

Step 1: Factoring x^{2} -16

x^{2} -16=x^{2} -4^{2} =(x+4)(x-4)

Step 2: Identifying the LCD, we get (x+4)(x-4)

Step 3: Rewriting the second fraction by multiplying x+4 on both top and bottom of second fraction so that we get the LCD.

\frac{x}{x^2-16}-\frac{3}{x-4}=\frac{x}{(x+4)(x-4)}   -\frac{3*(x+4)}{(x-4)*(x+4)} Step 4: Combine like terms since the denominators are same[tex] \frac{x-3(x+4)}{(x+4)(x-4)} =\frac{x-3x-12}{(x+4)(x-4)}=\frac{-2x-12}{(x+4)(x-4)} =\frac{-2(x+6)}{(x+4)(x-4)}

Conclusion:

In factored form the simplified expression =\frac{-2(x+6)}{(x+4)(x-4)}

In expanded form the simplified expression =\frac{-2x-12}{x^2-16}

8 0
3 years ago
emily is entering a bicycle race for charity her mothe pledges $0.40 for every 0.25 mile she bikes.if emily bikes 15 miles how m
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The answer is $24

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3 years ago
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storchak [24]

Answer:

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Step-by-step explanation:

A function is a set of relations without repetitive domains.

From the given relation, there are no repetitive domains. Thus our relation here is a function.

<u>Example</u><u> </u><u>of</u><u> </u><u>Function</u>

{(1,1), (2,2), (3,3), (4,4), (5,5)}

<u>Example</u><u> </u><u>of</u><u> </u><u>Non-Function</u>

{(1,1), (1,2), (2,3), (3,4), (4,5)}

From this relation, there are two repetitive domains which are 1's. Thus not making the relation a function.

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Answer:

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Step-by-step explanation:

hope this helps :)

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