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Rasek [7]
3 years ago
11

In Exercises 51−56, the letters a, b, and c represent nonzero constants. Solve the equation for x

Mathematics
2 answers:
labwork [276]3 years ago
8 0

▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓

x+a=\dfrac{3}{4}\\\\x=\dfrac{3}{4}-a\\\\x=\dfrac{3-4a}{4}

▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓

OverLord2011 [107]3 years ago
6 0

Answer:

x = ¾-a

Step-by-step explanation:

x + a = ¾

Subtract a from each side

x + a -a= ¾-a

x = ¾-a

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A cyclist rides her bike at a rate of 12 meters per second. What is this rate in kilometers per hour? How many kilometers will t
IgorLugansk [536]

Answer:

Part 1) The rate is 43.2\frac{km}{h}

Part 2) 129.6\ km

Step-by-step explanation:

Part 1)

we know that

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Remember that

1 hour=3,600 seconds

1 kilometer=1,000 meters

therefore

12\frac{m}{sec} =\frac{(12/1,000)}{(1/3,600)}=\frac{12*3,600}{1,000}=43.2\frac{km}{h}

Part 2) How many kilometers will the cyclist travel in 3 hours?

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3 0
3 years ago
X/x-2 + x-1/x+1= -1<br> Can someone walk me through the steps of this problem? Much appreciated. :)
stealth61 [152]

Answer:

x = 0 or x = 1

Step-by-step explanation:

Given equation:

\dfrac{x}{x-2}+\dfrac{x-1}{x+1}=-1

Multiply the two denominators to get the same denominator:

Multiply the numerator of the first fraction by the denominator of the second fraction to get the new numerator of the first fraction.

Multiply the numerator of the second fraction by the denominator of the first fraction to get the new numerator of the second fraction.

\implies \dfrac{x(x+1)}{(x-2)(x+1)}+\dfrac{(x-2)(x-1)}{(x-2)(x+1)}=-1

Add the numerators and keep the denominator so that there is now one fraction:

\implies \dfrac{x(x+1)+(x-2)(x-1)}{(x-2)(x+1)}=-1

Simplify the numerator by expanding the brackets:

\implies \dfrac{x^2+x+x^2-3x+2}{(x-2)(x+1)}=-1

\implies \dfrac{2x^2-2x+2}{(x-2)(x+1)}=-1

Simplify the denominator by expanding the brackets:

\implies \dfrac{2x^2-2x+2}{x^2-x-2}=-1

Multiply both sides by the denominator of the left side:

\implies 2x^2-2x+2=-1(x^2-x-2)

Simplify and move everything to the left side:

\implies 2x^2-2x+2=-x^2+x+2

\implies 3x^2-3x=0

Factor:

\implies 3x(x-1)=0

Therefore:

\implies 3x=0 \implies x=0

\implies x-1=0 \implies x=1

Therefore, x = 0 or x = 1

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2 years ago
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