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3 years ago

• 6x – 15y = 15 y=2 x - 1 5

Mathematics

1 answer:

yuradex [85]3 years ago

8 0

Answer:

Slope=

2.000

0.800

=0.400

x−intercept=

2 /5

=2.50000

y−intercept=

−5 /5

−1

=−1.00000

Step-by-step explanation:

STEP

Pulling out like terms

1.1 Pull out like factors :

6x - 15y - 15 = 3 • (2x - 5y - 5)

Equation at the end of step

STEP

Equations which are never true

2.1 Solve : 3 = 0

This equation has no solution.

A a non-zero constant never equals zero.

Equation of a Straight Line

2.2 Solve 2x-5y-5 = 0

Tiger recognizes that we have here an equation of a straight line. Such an equation is usually written y=mx+b ("y=mx+c" in the UK).

"y=mx+b" is the formula of a straight line drawn on Cartesian coordinate system in which "y" is the vertical axis and "x" the horizontal axis.

In this formula :

y tells us how far up the line goes

x tells us how far along

m is the Slope or Gradient i.e. how steep the line is

b is the Y-intercept i.e. where the line crosses the Y axis

The X and Y intercepts and the Slope are called the line properties. We shall now graph the line 2x-5y-5 = 0 and calculate its properties

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If f(x) = |(x2 − 9)(x2 + 1)|, how many numbers in the interval [−1, 1] satisfy the conclusion of the mean value theorem?

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Answer:

only one number c=0 in the interval [-1,1]

Step-by-step explanation:

Given : Function $f(x) = |(x^2-9)(x^2 + 1)|$ in the interval [-1,1]

To find : How many numbers in the interval [−1, 1] satisfy the conclusion of the mean value theorem.

Mean value theorem : If f is a continuous function on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a point 'c' in (a,b) such that $f'(c)=\frac{f(b)-f(a)}{b-a}$

Solution : f(x) is a function that satisfies all of the following :

1) f(x) is continuous on the closed interval [-1,1]

$\lim_{x\to a} f(x)=f(a)$

2) f(x) is differentiable on the open interval (-1,1)

Then there is a number c such that $f'(c)=\frac{f(b)-f(a)}{b-a}$

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$f(b)=f(1) = |(1^2-9)(1^2 + 1)|=|(8)(2)|=16$

$f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{16-16}{2}=0$

$f'(c)=0$ ........[1]

Now, we find f'(x)

$f(x) = |(x^2-9)(x^2 + 1)|$

$f(x) =x^4-8x^2-9$

Differentiating w.r.t x

$f'(x) =4x^3-16x$

In place of x we put x=c

$f'(c) =4c^3-16c$

$f'(c) =4c^3-16c=0$ (by [1], f'(c)=0)

$4c(c^2-4)=0$

$4c=0,c^2-4=0$

either c=0 or $c^2-4=0\rightarrow c=\pm2$

we cannot take $c=\pm2$ because they don't lie in the interval [-1,1]

Therefore, there is only one number c=0 which lie in interval [-1,1] and satisfying the conclusion of the mean value theorem.

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