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mihalych1998 [28]

3 years ago

5

Can you help me with try these B?

1 answer:

Softa [21]3 years ago

6 0

A. expanded, you get XxXxXxXxXxXxXxX over XxXxXxXxX
b.XxXxX or X³
c. I would just cancel out the 5 Xs on the top line against the 5 Xs on the bottom, leaving X³.

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13 = 2m + 5<br><br> whats m?

timama [110]

4 is the correct answer

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2 years ago

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spayn [35]

Answer:

see below

Step-by-step explanation:

whenever there is negative number being squared, you need to add the brackets so that you inside the - 1 in front of the number

so this step is wrong

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so it should be

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2 years ago

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Is 9(-5)^x exponential?

olga2289 [7]

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Step-by-step explanation:

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2 years ago

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ICE Princess25 [194]

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2 years ago

Find a closed-form solution to the integral equation y(x) = 3 + Z x e dt ty(t) , x > 0. In other words, express y(x) as a fun

MrMuchimi

Answer:

$y{x} = \sqrt{7+2Inx}$

Step-by-step explanation:

$y(x)= 3 + \int\limits^x_e {dx}/ \, ty(t) , x>0}$

Let say; By y(x)= y(e)

we have;

$y(e)= 3 + \int\limits^e_e {dt}/ \, ty= 3+0$

Using Fundamental Theorem of Calculus and differentiating by Lebiniz Rule:

$y^{1} (x) = 0 + 1/ xy$

$y^{1} = 1/xy$

dy/dx = 1/xy

$\int\limits {y} \, dxy = \int\limits \, dx/x$

$y^{2}/2 Inx + C$

RECALL: y(e) = 3

$(3)^{2} / 2 = In (e) + C$

$\frac{9}{2} =In(e)+C$

$\frac{9}{2} - 1 = C$

$\frac{7}{2} = C$

$y^{2} / 2 = In x +C$

$y^{2} / 2 = In x +7/2$

MULTIPLYING BOTH SIDE BY 2 , TO ELIMINATE THE DENOMINATOR, WE HAVE;

$y^{2} = {7+2Inx}$

$y{x} = \sqrt{7+2Inx}$

8 0

3 years ago