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Vanyuwa [196]
3 years ago
14

How do you find the slope of x^2 at x = 5? please hurry

Mathematics
1 answer:
Lorico [155]3 years ago
5 0

The slope of any function is its first derivative.

If  f(x)=x^2, then  f'(x)= 2x .

At  x=5,  slope = f'(x) = 2(5) = <em>10</em> .
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Read 2 more answers
2. Which of the following equations are perpendicular to 2y = -3x + 1 I. II. III. y=-x-1 - 2x + 3y = -5 2x + 3y = 2 (A) I only (
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The equation in given as ;

2y = -3x + 1

This can be written as ;

y= -3/2 x + 1/2

This means the equation has a gradient of -3/2

Let this slope , be , ---------m1

For perperdicular lines , the product of their slopes = -1 .This means if the other line has a gradient of m2 then : m1 * m2 = -1

So from the answers :

i) y= 2/3 x - 1 the slope is 2/3

m2 = 2/3

m1 * m2 = -1 -------check the if this is true by using the two values of gradient as;

-3/2 * 2/3 = - 1 ------ This is true-----equation i

II.

-2x + 3y = -5

3y = 2x -5

y= 2/3 x -5/3 -----m2 here is 2/3

m1*m2 = -1

-3/2 * 2/3 = -1 -----this is true , so ----equation ii

iii)

2x + 3y = 2

3y = -2x + 2

y= -2/3 x + 2/3 -----m2 = -2/3

m1*m2 = -1

-3/2 * -2/3 = 1 -----this is not true,,,equation iii is not perpendicular to our equation.

so, equation i and ii are perpendicular to our equation .

Answer : B i and ii only

4 0
1 year ago
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