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3 years ago

Juanita studied 3/2 hours on Saturday and 5/4 hours on Sunday. How many total hours did she study those two days?

Mathematics

1 answer:

lord [1]3 years ago

6 0

Answer:

11/4

Step-by-step explanation:

3/2 can be re-written as 6/4. they now have the same denominator. 6/4 + 5/4 = 11/4

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The value of c is equal to _____.

DENIUS [597]

For this case we have that by definition, the Pythagorean theorem states that:

$c = \sqrt {a ^ 2 + b ^ 2}$

Where:

a, b: Are the legs

c: It is the hypotenuse

According to the figure we have to:

$a = 20\\b = 21$

Substituting in the formula we have:

$c = \sqrt {(20) ^ 2 + (21) ^ 2}\\c = \sqrt {400 + 441}\\c = \sqrt {841}\\c = 29$

Answer:

Option D

8 0

3 years ago

What is the value of F(5) in the function below?

Sever21 [200]

Answer:

Step-by-step explanation:

F(x) = 3^x

then

F(5) = 3^5 = 3×3×3×3×3 =243

6 0

4 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

I need help finding the circumference of F

Arada [10]

So your circumference would be <span>34.557519189488 cm</span>. hope that helped

3 0

3 years ago

Can u please help me and show me how u got that answers.

GREYUIT [131]

C. 20% because 15 goes into 75 5 times, and to find the percentage you would divide 100 by that number, 5, to get 20. So he has gone 20%

6 0

3 years ago

Read 2 more answers