Ask question

Ask question

All categories

4 years ago

14

Number 8. Find the mean please show your work

1 answer:

lukranit [14]4 years ago

7 0

Find the mean of the number of seats? Or the amount of cars?

You might be interested in

The sales tax for an item was $32.80 and it cost $410 before tax

Darya [45]

Answer:

do you have any answers to choose from?

8 0

3 years ago

What is the distance between (9,5) and (-2,2)

nikitadnepr [17]

Answer:

10 units

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

What is the quotient of 65,610 ÷ 18?

ankoles [38]

The answer to this is 3645

5 0

3 years ago

Read 2 more answers

Drawing Conclusions

maksim [4K]

Answer:

Some of the Exponents = -2 that is true, not 2.

Step-by-step explanation:

Let's check one at a time.

(a)The 6 without an exponent is equivalent to the 6 having a 0 exponent.

$6^{0} =1$ and $6^{1}$ = 6 (no exponent. 6 $\neq$ 1 therefore this statement is False.

(b)The sum of the exponents is -2.

let's check , if the base is same we can add the exponents that is the exponent rule.(well established).

if we add exponents in the given expression we get.

$6^{1} 6^{0} 6^{3} =6^{1+0+(-3)} =6^{-2}$ , therefore we can see that the sum of the exponents = -2 this is true.

(c) An equivalent expression is 65.6-7, lets evaluate our above expression, it is equal to $\frac{1}{36}$ which we can see that $\frac{1}{36} \neq 65.6-7$ ,therefore this statement is false as well.

3 0

3 years ago

Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2

Verdich [7]

Answer:

(a) $f'(x)=-\frac{2}{x^3}$

(b) $y=-0.25x+0.75$

Step-by-step explanation:

The given function is

$f(x)=\frac{1}{x^2}$ .... (1)

According to the first principle of the derivative,

$f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}$

Cancel out common factors.

$f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}$

By applying limit, we get

$f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}$

$f'(x)=\frac{-2x)}{x^4}$

$f'(x)=\frac{-2)}{x^3}$ .... (2)

Therefore $f'(x)=-\frac{2}{x^3}$ .

(b)

Put x=2, to find the y-coordinate of point of tangency.

$f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25$

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

$m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}$

Substitute x=2 in equation 2.

$f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25$

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

$y-y_1=m(x-x_1)$

$y-0.25=-0.25(x-2)$

$y-0.25=-0.25x+0.5$

$y=-0.25x+0.5+0.25$

$y=-0.25x+0.75$

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

5 0

4 years ago