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2 years ago

14

PLs answer the question in the picture. I will give a thanks, 5 stars, and BRAINLIEST for a good and applicable answer.

2 answers:

____ [38]2 years ago

8 0

The big angle is < BAD

Shkiper50 [21]2 years ago

3 0

Answer:

B

Step-by-step explanation:

its the answer

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What is the solution to 2x+1=-7x+19

joja [24]

The answer is x equals 2

7 0

3 years ago

Yellow Taxi Cabs charges a $7 flat rate in addition to $2.25 per mile. Kate has $100 to spend on transportation. How many miles

enyata [817]

Kate can travel 41.33 miles without exceeding her limit. This problem can be solved by using y = 2.25x + 7 linear equation with the "y" variable as the total cost that Kate must pay after she has traveled with the cab and the "x" variable as Kate's traveling distance. The equation has 7 for its constant value which is the $7 flat rate. We will find 41.33 miles as the traveling distance if we substituted the total cost with 100, which is the maximum amount that can be paid by Kate for the traveling purpose.

7 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

3 years ago

Somebody helpppppppppppoo

Marizza181 [45]

Answer:

40 degrees : )

4 0

3 years ago

Which equation has the correct sign on the product?

Alik [6]

Answer:

B

Step-by-step explanation:

(+)x(-)=(-)

(-)x(-)=(+)

8 0

3 years ago