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Dimas [21]
2 years ago
8

Solve each of the following exponential equations using common bases:

Mathematics
1 answer:
pantera1 [17]2 years ago
6 0

Answer:

x=19/6, x=-9/13, x=33/4

Step-by-step explanation:

1) 2^(6x+10)=2^(12x-9), 6x+10=12x-9, x=19/6

2) 3^(4x+6)=3^(-9x-3), 4x+6=-9x-3, x=-9/13

3) 2^(x+6)*2^(3x-27)=2^(6), 4x-27=6, x=33/4

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Answer:  < -4/5,  3/5>

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======================================================

Explanation:

Draw an xy grid and plot the point (-4,3) on it. Draw a segment from the origin to this point. Then draw a vertical line until reaching the x axis. See the diagram below.

We have a right triangle with legs of 4 and 3. The hypotenuse is \sqrt{4^2+3^2} = \sqrt{16+9} = \sqrt{25} = 5 through use of the pythagorean theorem.

We have a 3-4-5 right triangle.

Therefore, the vector is 5 units long. This is the magnitude of the vector.

Divide each component by the magnitude so that the resulting vector is a unit vector pointing in this same direction.

Therefore, we go from < -4, 3 > to < -4/5,  3/5 >

This is equivalent to < -0.8, 0.6 > since -4/5 = -0.8 and 3/5 = 0.6

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2 years ago
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Marina86 [1]
Answer:\frac{6}{y^2-xy}-\frac{6}{x^2-xy}=\frac{6(x+y)}{xy(y-x)}

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Combining the fractions, we have:

\begin{gathered} \frac{6(x^2-xy)-6(y^2-xy)}{(x^2-xy)(y^2-xy)} \\  \\ =\frac{6x^2-6xy-6y^2+6xy}{(x^2-xy)(y^2-xy)} \\  \\ =\frac{6x^2-6y^2}{x(x-y).y(y-x)} \\  \\ =\frac{6(x-y)(x+y)}{-xy(x-y)^2} \\  \\ =\frac{-6(x+y)}{xy(x-y)} \\  \\ =\frac{6(x+y)}{xy(y-x)} \end{gathered}

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