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oksian1 [2.3K]
3 years ago
14

What is the equation for the line in slope-intercept form? will give brainliest

Mathematics
1 answer:
andriy [413]3 years ago
7 0

Answer:

<em>y </em>= 8<em>x</em>

Step-by-step explanation:

The slope is 8, the y-intercept is 0. The slope-intercept form is written:

<em>y</em> = <em>mx</em> +<em> b</em>

So this situation would be written <em>y =  </em>8<em>x </em>+ 0 or just <em>y</em> = 8<em>x.</em>

Hope it helps!

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Step-by-step explanation:

This differential equation has separable variable and can be solved by integration. First derivative is now obtained:

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f' = \int {\left(x-\frac{3}{2}\right) } \, dx

f' = \int {x} \, dx -\frac{3}{2}\int \, dx

f' = \frac{1}{2}\cdot x^{2} - \frac{3}{2}\cdot x + C, where C is the integration constant.

The integration constant can be found by using the initial condition for the first derivative (f'(4) = 1):

1 = \frac{1}{2}\cdot 4^{2} - \frac{3}{2}\cdot (4) + C

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y = \int {\left(\frac{1}{2}\cdot x^{2}-\frac{3}{2}\cdot x -1  \right)} \, dx

y = \frac{1}{2}\int {x^{2}} \, dx - \frac{3}{2}\int {x} \, dx - \int \, dx

y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x + C

C = 0 - \frac{1}{6}\cdot 0^{3} + \frac{3}{4}\cdot 0^{2} + 0

C = 0

Hence, the particular solution of the differential equation is y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x.

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