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3 years ago

9

MULTIPLE CHOICE QUESTIONS

1 answer:

IgorC [24]3 years ago

5 0

Answer:

C due to distrubtive properties

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A train stops at Atlanta, then at Bayville, and then at Colleyville. The

Nezavi [6.7K]

Answer:

See Explanation

Step-by-step explanation:

<em>The question is incomplete as what is required of the question is not stated.</em>

<em>However, since the question is only limited to distance, a likely question could be to calculate the distance from Bayville to Colleyville.</em>

Represent the distance from Atlanta to Colleyville with AC

Represent the distance from Atlanta to Bayville with AB

Represent the distance from Bayville to Colleyville with BC

So, we have that:

$AC = 25.5\ miles$

$AB = 10\frac{2}{3}\ miles$

The relationship between AB, AC and BC is:

$AC = AB + BC$

Make BC the subject of formula:

$BC = AC - AB$

$BC = 25.5 - 10\frac{2}{3}$

Convert fraction to decimal

$BC = 25.5 - 10.7$

$BC = 14.8$

<em>Hence, the distance from Bayville to Colleyville is 14.8 miles</em>

3 0

3 years ago

Write the ratio of corresponding sides for the similar triangles and reduce the ratio to lowest terms.

Schach [20]

Answer:

ANSWER is B. look at the sides and simplified.

5 0

3 years ago

Chad and Karla are both 42 years old they open an account at the laxmi savings bank with the hope that it will gain enough inter

dexar [7]

Idk

Step-by-step explanation:

your head is so big that you need brainly to answer questions so

4 0

3 years ago

What is the circumference of this circle, in millimeters? Use for .

Mrrafil [7]

Answer:

Step-by-step explanation:

C= d*pi

49 x 2 = 98

98 x 3.14= 307.72

Since 308 is the closest to 307.72 it's the most accurate answer.

6 0

3 years ago

Read 2 more answers

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

3 years ago