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snow_tiger [21]
3 years ago
6

Need help with a simple problem 2^x=3^x+1

Mathematics
1 answer:
KiRa [710]3 years ago
4 0
For x\ \textgreater \ 0, 2^x\ \textless \ 3^x, all the more 2^x\ \textless \ 3^x+1, therefore no solution.

For x=0, we have 2^0=3^0+1 \Rightarrow 1=1+1 \Rightarrow 1=2. Obviously that' false.

For x\ \textless \ 0 both 2^x and 3^x, where 2^x\ \textgreater \ 3^x, belong to the range (0,1), then 3^x+1 belong to the range (1,2).
Those ranges don't have a common part, therefore, again, no solution.

So, the equation 2^x=3^x+1 doesn't have a solution.

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A

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Someone please convert this equation to slope intercept form: 2x - 3y = 6?
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All you have to do is rearrange the equation:

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8 0
4 years ago
At a cherry farm, the cost per pound of cherries depends on the amount of cherries bought. For an amount up to 100 pounds, the c
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they would have saved 35 dollars

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3 0
3 years ago
Type your answer and then click or tap Done.<br> Simplify -(x-2y) - y.
AlladinOne [14]

Answer:

Step-by-step explanation:

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Step by step solution :

Step  1  :

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4 0
3 years ago
Read 2 more answers
Help pls- <br><br> I dont know it T^T
docker41 [41]

Answer:

Ans: H and A

Step-by-step explanation:

By the identity:

1-(cos(x))cos^{2}(x)+ sin^{2}(x)=1\\cos^{2}(x)=1- sin^{2}(x)\\\\cos(x)=\sqrt{1- sin^{2}(x)} \\sin(x)=\sqrt{1- cos^{2}(x)}\\\\\sqrt{1- cos^{2}(x)}/sinx=1\\\sqrt{1- sin^{2}(x)}/cosx=1\\\sqrt{1- cos^{2}(x)}/sinx+\sqrt{1- sin^{2}(x)}/cosx=2\\\\\\

Ans: H

For second quesiton, the transformation of a graph is that:

f(x) + k means vertical translation up by k units

Two curve have the same maximum value, so the graph doesn't translate upwards or downwards, thus b=0

Ans: A

7 0
3 years ago
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