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3 years ago

Plz help.......................'

Mathematics

1 answer:

tekilochka [14]3 years ago

5 0

It is the second one.
r2/r2 cancels out.
s5/s2=s3
t4/t5=t^-1 which is also 1/t.
If you combine the two terms you get s3/t which is the second one.

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find the equations of the tangents to the curve y= x(x-1)(x+2) at the points where the curve cuts the x axis

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First of all, we compute the points of interest, i.e. the points where the curve cuts the x axis: since the expression is already factored, we have

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Which means that the roots are

$x=0\ \lor\ x=1\ \lor\ x=-2$

Next, we can expand the function definition:

$y = x(x-1)(x+2) = x^3 + x^2 - 2x$

In this form, it is much easier to compute the derivative:

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$y'(-2) = 6,\quad y'(0)=-2,\quad y'(1)=3$

This means that we are looking for the equations of three lines, of which we know a point and the slope. The equation

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is what we need. The three lines are:

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3 years ago

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enot [183]

Hello!

To solve this, first perform the opposite operation for the last operation (on the left side) on both sides. The last operation of the left side is squaring. Therefore, square root both sides.

$(y - 9)^{2} = 64$

$y - 9 = ±\sqrt{64}$

$y - 9 = ±8$

Please note that you must include ±. This is because the square root of 64 can be either positive or negative, as a square of either a positive or negative number is positive.

Now, add 9 to both sides.

$y = 9±8$