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Vikki [24]
3 years ago
13

Solve 3^(2x) = 7^(x_1).

Mathematics
2 answers:
uranmaximum [27]3 years ago
8 0
Applying log to both sides of equation, we get:

<span>2x <span>log 3</span>=<span>(x−1) </span><span>log 7</span></span>

x=−0.12915

<span><span>
</span></span>

bulgar [2K]3 years ago
4 0
<span>x=1−<span><span>log9</span><span>log7</span></span>=−0.12915</span><span>, nearly.</span>
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Two angle measuresof this triangle are 60° which type triangles is it?
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Answer:

An equilateral triangle has all three sides equal, and all three interior angles equal, too. In this case, each interior angle of an equilateral triangle is 60 degrees.

Step-by-step explanation:

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2 years ago
Daniel volunteers on the weekend at the Central Library. As a school project, he decides to record how many people visit the lib
liq [111]

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500

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hope it helps you

tic tac toe

7 0
3 years ago
Maria stated that 5/6 is between 4/5 and 6/7
Vlada [557]

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It's true, if that's what your asking.

Step-by-step explanation:

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3 0
3 years ago
Kim drives 156 miles from rotherham to London.She has an average speed of 60 miles per hour.
kicyunya [14]
Currently, you would ask "How long does it take to get from Rotherham to London?" You would do 156/60, which would be 2.6 hours. 2.6 hours is 156 minutes (2 hours, 36 minutes).
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5 0
3 years ago
Linda goes water-skiing one sunny afternoon. After skiing for 15 min, she signals to the driver of the boat to take her back to
trasher [3.6K]
60 = a * (-30)^2
a = 1/15
So y = (1/15)x^2


abc)
The derivative of this function is 2x/15. This is the slope of a tangent at that point.
If Linda lets go at some point along the parabola with coordinates (t, t^2 / 15), then she will travel along a line that was TANGENT to the parabola at that point.
Since that line has slope 2t/15, we can determine equation of line using point-slope formula:
y = m(x-x0) + y0
y = 2t/15 * (x - t) + (1/15)t^2
Plug in the x-coordinate "t" that was given for any point.


d)
We are looking for some x-coordinate "t" of a point on the parabola that holds the tangent line that passes through the dock at point (30, 30).
So, use our equation for a general tangent picked at point (t, t^2 / 15):
y = 2t/15 * (x - t) + (1/15)t^2
And plug in the condition that it must satisfy x=30, y=30.
30 = 2t/15 * (30 - t) + (1/15)t^2
t = 30 ± 2√15 = 8.79 or 51.21
The larger solution does in fact work for a tangent that passes through the dock, but it's not important for us because she would have to travel in reverse to get to the dock from that point.
So the only solution is she needs to let go x = 8.79 m east and y = 5.15 m north of the vertex.
4 0
3 years ago
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