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3 years ago

12

Use the number line to answer the question. Each tick represents 1.

1 answer:

snow_lady [41]3 years ago

6 0

s because before 0 is negative numbers

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A = {a, b, c, d}

Helen [10]

Answer:

Johnny is wrong.

$A \cup B = \left\{a,b,c,d,e,f,g,h\right\}$

Step-by-step explanation:

Johnny is wrong.

A better definition would be: $A \cup B$ is the set of all elements that belong to at least one A or B.

So, the elements that belong to both A and B, like c and d in this exercise, also belong to $A \cup B$ .

So:

$A \cup B=\left\{a,b,c,d,e,f,g,h\right\}$

4 0

3 years ago

Is this correct? helppp

Marta_Voda [28]

Answer:

Yes its correct the unhighlighted is complement

8 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

4 years ago

Please help me fast please , i really dont understand .

Sever21 [200]

Answer:

y= 1/3x - 5/2 or D

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

HELPPPPPPPPPPPPPPPPPPPPPPP

nadya68 [22]

Answer:

that is a lot

Step-by-step explanation:

6 0

3 years ago