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iren [92.7K]
3 years ago
6

Help ASAP please!!!!!!

Mathematics
1 answer:
AURORKA [14]3 years ago
8 0

Answer:

x5 {36}^{?65}  \\

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SUPER EASY 19 POINTS
Helga [31]

Answer:

I mean I’m not to sure on how you would show work for it but the answer would be 0.123 hope this helps

Step-by-step explanation:

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the function of y=log(x) is translated 1 unit right and 2 units down. which is the graph of the translated function
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The answer is B. because its B.
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What us the ratio of 8 pennies and 20 dimes
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If the legs of a right triangle are x, (2x - 1) and its hypotenuse is (2x + 1), what is the value of x?​
Ber [7]

Answer:

x=8

Step-by-step explanation:

(2x + 1)² = x² + (2x - 1)² = x² + (2x - 1)(2x - 1)

(2x + 1)(2x + 1) = x² + 4x²-2x-2x+1

4x²+2x+2x+1 = 5x²-4x+1

4x²+4x+1 = 5x²-4x+1

0= 5x²-4x²-4x-4x+1-1

0= x²-8x

x²-8x=0

x²=8x

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3 0
3 years ago
(b) dy/dx = (x - y+ 1)^2
Elanso [62]

Substitute v(x)=x-y(x)+1, so that

\dfrac{\mathrm dv}{\mathrm dx}=1-\dfrac{\mathrm dy}{\mathrm dx}

Then the resulting ODE in v(x) is separable, with

1-\dfrac{\mathrm dv}{\mathrm dx}=v^2\implies\dfrac{\mathrm dv}{1-v^2}=\mathrm dx

On the left, we can split into partial fractions:

\dfrac12\left(\dfrac1{1-v}+\dfrac1{1+v}\right)\mathrm dv=\mathrm dx

Integrating both sides gives

\dfrac{\ln|1-v|+\ln|1+v|}2=x+C

\dfrac12\ln|1-v^2|=x+C

1-v^2=e^{2x+C}

v=\pm\sqrt{1-Ce^{2x}}

Now solve for y(x):

x-y+1=\pm\sqrt{1-Ce^{2x}}

\boxed{y=x+1\pm\sqrt{1-Ce^{2x}}}

3 0
3 years ago
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