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Stolb23 [73]
3 years ago
12

8^x+3=16^x−1 solve the exponential equation ...?

Mathematics
1 answer:
Juli2301 [7.4K]3 years ago
6 0
Let us write down the equation clearly and in a proper manner. This will help in solving the equation very easily.

8^(x + 3) = 16^(x - 1)
(2^3)^(x + 3) = (2^4)^(x - 1)
(2)^3(x + 3) = (2)^4(x - 1)
Then
3(x + 3) = 4(x - 1)
3x + 9 = 4x - 4
4x - 3x = 9 + 4
x = 13

I hope the procedure is clear enough for you to understand. I also hope that this is the answer that has actually come to your desired help.
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Answer:

the slope is 6/7 the y-intercept is (0, -\frac{15}{7})

Step-by-step explanation:

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Lydia graphed ADEF at the coordinates D (-2,-1), E (-2, 2), and F (0,0). She thinks ADEF is a right triangle. Is Lydia's asserti
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The true statement is (c) No; the slopes of segment EF and segment DF are not opposite reciprocals.

<h3>Right triangles </h3>

Right triangles have a pair of perpendicular lines

Coordinates

The coordinates are given as:

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  • E = (-2,2)
  • F = (0,0)

<h3>Slopes</h3>

Start by calculating the slopes of lines DF and EF using:

m = \frac{y_2 -y_1}{x_2 -x_1}

So, we have:

m_{DF} = \frac{0 + 1}{0 +2}

m_{DF} = \frac{1}{2}

Also, we have:

m_{EF} = \frac{0 -2 }{0+2}

m_{EF} = \frac{-2 }{2}

m_{EF} = -1

For the triangle to be a right triangle, then the calculated slopes must be opposite reciprocals.

i.e.

m_1 = -\frac{1}{m_2}

By comparison, the slopes of both lines are not opposite reciprocals.

Hence, the true statement is (c)

Read more about right triangles at:

brainly.com/question/17972372

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Find x: 1) 2x= 1/8 2)3x multiplied by 9= 81
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Heres your answer girlfriend ♥️♥️

Have you followed me when I answered your last question???

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3 years ago
Why the derivative of (x^2/a^2) = (2x/a^2)? ​
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I assume you're referring to a function,

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where <em>a</em> is some unknown constant. By definition of the derivative,

\displaystyle f'(x) = \lim_{h\to0}{f(x+h)-f(x)}h

Then

\displaystyle f'(x) = \lim_{h\to0}{\frac{(x+h)^2}{a^2}-\frac{x^2}{a^2}}h \\\\ f'(x) = \frac1{a^2} \lim_{h\to0}{(x+h)^2-x^2}h \\\\ f'(x) = \frac1{a^2} \lim_{h\to0}{(x^2+2xh+h^2)-x^2}h \\\\ f'(x) = \frac1{a^2} \lim_{h\to0}{2xh+h^2}h \\\\ f'(x) = \frac1{a^2} \lim_{h\to0}(2x+h) = \boxed{\frac{2x}{a^2}}

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