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stealth61 [152]
3 years ago
11

3(x+2)=9(6-x) Answer please and show your work :D

Mathematics
1 answer:
Nookie1986 [14]3 years ago
6 0

Answer: x=4

Step-by-step explanation: 3x+6= 54-9x

6= 54 -12x

-48=-12x

x=4

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6 0
2 years ago
Which expression is equivalent to the following complex fraction? StartFraction 1 Over x EndFraction minus StartFraction 1 Over
Darina [25.2K]

Answer:

\frac{y-x}{x+y}

Step-by-step explanation:

We are given that fraction

\frac{\frac{1}{x}-\frac{1}{y}}{\frac{1}{x}+\frac{1}{y}}

We have to find the expression which is equivalent to given  fraction .

\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}

\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}

Substitute the values  then, we get

\frac{\frac{y-x}{xy}}{\frac{y+x}{xy}}

We know that

\frac{\frac{a}{b}}{\frac{x}{y}}=\frac{a}{b}\times \frac{y}{x}

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6 0
3 years ago
Read 2 more answers
Could someone help me, please?
k0ka [10]

We can rewrite the expression under the radical as

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then taking the fourth root, we get

\sqrt[4]{\left(\dfrac32a^2b^3c^4\right)^4}=\left|\dfrac32a^2b^3c^4\right|

Why the absolute value? It's for the same reason that

\sqrt{x^2}=|x|

since both (-x)^2 and x^2 return the same number x^2, and |x| captures both possibilities. From here, we have

\left|\dfrac32a^2b^3c^4\right|=\left|\dfrac32\right|\left|a^2\right|\left|b^3\right|\left|c^4\right|

The absolute values disappear on all but the b term because all of \dfrac32, a^2 and c^4 are positive, while b^3 could potentially be negative. So we end up with

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3 0
3 years ago
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VMariaS [17]

Answer :SOLVING EQUATIONS CONTAINING ABSOLUTE VALUE(S)

Step 1: Isolate the absolute value expression.

Step2: Set the quantity inside the absolute value notation equal to + and - the quantity on the other side of the equation.

Step 3: Solve for the unknown in both equations.

Step 4: Check your answer analytically or graphically.

Step-by-step explanation:

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NeTakaya
All of it is 8 because sides are all equal
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