I need to find the slope and y-intercept of 5x-2y=3. I think the slope is 5/2 and the y-intercept is -3/2???? Is this correct?

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I need to find the slope and y-intercept of 5x-2y=3. I think the slope is 5/2 and the y-intercept is -3/2???? Is this correct?

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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yes it is .....can we work on a partial diffferential problem?
sorry don't have any problems like that right now.
i do have ...and need someone to workout on it...i want to match my answer with it

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slope and intercept on y axis are correct
whats the second order partial derivative when function f(x,y) =Sin xy. both wrt x and y
Thinker: The derivative of sin(f(x,y)) w.r.t. x would be cos(f(x,y)) * df(x,y)/dx. In other words, take the derivative of cosine, then take the derivative of what's inside the cosine. Applying that formula, the d(sin(xy))/dx = cos(xy)y. Does that help?
i have differentiated w.r.t. x and got −y2Sinxy but to find 2nd order part. derv. i have differentiated this w.r.t. y and i have got −xySinxy−cosxy is this correct?
I'm not quite sure what '2nd order partial derivative' is (and I'm confused by your notation). But if you take the second derivative of sin(xy) w.r.t. x, you get -y^2*sin(xy) (negative y squared times the sine of x times y). The second deriv wrt y is -x^2*sin(xy). You can also take partial derivatives wrt x, then the partial derivative of that wrt to y.
yes i have got 1st derv(after diff wrt x, say eq 1) as wat you got, now i need to differentiate this eq. wrt y...i.e. eq 1 wrt 'y'
Then you'll need to use the product rule on cos(xy)y. For the product rule, u=cos(xy), and v=y. The product rule is (uv)'=u'v+uv' (the derivative of u times v equals the derivative of u times y plus u times the derivative of v. I think you have enough to complete the problem now.

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