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g100num [7]
3 years ago
6

Geneva rode her bike a total of 2 1/2 miles from her house to school. First she rode 4/5 mile from her house to the park. Then s

he rode 1/5 mile from the park to her friend’s house. Finally she rode the rest of the way to her school. How many miles did she ride from her friend's house to school?
Mathematics
1 answer:
Sonja [21]3 years ago
8 0
Add 4/5 and 1/5 to make 5/5 or 1 mile the subtract 1 from 2 1/2 to equal 1 1/2 miles
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nydimaria [60]
Answer is 5040
10×9×8×7=5040
5 0
3 years ago
a car uses 57 liters of gasoline for every 644 kilometers traveled. it takes 8 hours to use all the gasoline in the tank. what i
devlian [24]

Answer:

80.5 kilometers/h

Step-by-step explanation:

S=80.5km/h

D644

T8 hours

8 0
3 years ago
In 2013, voyager 1 traveled 1,468,800 kilometers each day. Whats that in scientific notation
n200080 [17]

Answer: 1.4688 x 10^6

Step-by-step explanation: Number has to be in between 1 and 10. Moved the decimal 6 places so that is your exponent

8 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
find the missing side. round to the nearest tenth. use pythagorean theorem to find the third side length​
emmasim [6.3K]

Answer:

x is tangent of 39 = x/ 19, tangent of 39 is 3. 615...... then 19 multiply by 3.615....., x would get 68. 68

Step-by-step explanation:

tangent is opposite over adjacent then put all your numbers in to the formula.

4 0
3 years ago
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