Can someone explain

When driving in England, Brian drove through several speed cameras. On one drive, Brian left his hotel and after 7 minutes he wa

DedPeter [7]

Solution :

The formula for the linearization of a function $f(x)$ at a point $x$ = a is given as

$L(x)=f(a)+(x-a)f'(a)$

Assuming the time is t and the distance travelled is $f(t)$ , that makes the speed as $f'(t)$ .

So substituting them in the linearization formula,

A. At t = 7 minutes

f(7) = 2.5 km

f'(7) = 0.5 kpm

∴ $L_7(t)=f(7)+(t-7)f'(7)$

$=2.5+(t-7)0.5$

$=2.5+0.5t-3.5$

$=0.5t-1$

B. At t = 18 minutes

f(18) = 14.8 km

f'(18) = 0.8 kpm

∴ $L_{18}(t)=f(18)+(t-18)f'(18)$

$=14.8+(t-18)0.8$

$=14.8+0.8t-14.4$

$=0.8t-0.4$

C. Substituting the value of t as 14 in both the linearization to determine the position at 14 minutes, we get

$L_7(14)=0.5(14)-1.0$

= 7 - 1

= 6 km

$L_{18}(14)=0.8(14)+0.4$

= 11.2 + 0.4

= 11.6 km

D. According to the linearization at 7, the distance travelled between the 7 minutes and 14 minutes is = 6 km - 2.5 km

= 3.5 km

And between the 14 minutes and 18 minutes is = 14.8 km - 6 km

= 8.8 km

This is an average speed of 0.5 kpm in the first interval and an average speed of 2.2 kpm.

Now, according to the linearization of 18, the distance travelled between the 7 minutes and the 14 minutes is = 11.6 km - 2.5 km

= 9.1 km

And between 14 minutes and 18 minutes is = 14.8 km - 11.6 km

= 3.2 km

This gives an average speed of 1.3 kpm in the first interval and 0.8 kpm in the second interval.

Therefore, the second approximation is the better one since the average speed are closer to the actual readings in the second linearization.

4 0

3 years ago

How do i solve this type of rational equation (x^2-16)/(4-x)

babunello [35]

I thinks it’s -x-4 I’m not sure

8 0

3 years ago

Which of the following options represent the three undefined terms in geometry

hoa [83]

What are the options?

5 0

3 years ago

What is the factored form of the function f(x)=x^3+8x^2+5x−50?

Inga [223]

Answer:

the factors of f(x)=x^3+8x^2+5x-50 are (x-2)(x+5)(x+5)

Step-by-step explanation:

We need to factorise the function $f(x)=x^3+8x^2+5x-50$

If a number is a factor of this function than it must be completely divisible by last co-efficient. Our last co-efficient is -50

Checking few numbers:

$f(1)=(1)^{3}+8(1)^2+5(1)-50\\f(1)=1+8+5-50\\f(1)=-32\\Now \ putting \ x= 2 \\f(2)=(2)^{3}+8(2)^2+5(2)-50\\f(2)=8+8(4)+10-50\\f(2)=8+32+10-50\\f(2)=0$

So, f(2)=0 which means x-2 is a factor of the given function. Now we will perform long division of $x^3+8x^2+5x-50$ by (x-2) to find other factors

The long division is shown in figure attached.

After long division we get: $x^2+10x+25$

The equation $x^2+10x+25$ can be further simplified as: (x+5)(x+5) or (x+5)^2

So, the factors of f(x)=x^3+8x^2+5x-50 are (x-2)(x+5)(x+5)

3 0

3 years ago

You find one quarter, two dimes, and two nickels.

vitfil [10]

Answer:

a: 0.55

b:11/20

Step-by-step explanation:

hope this helps

5 0

2 years ago

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