All categories

1 year ago

7-18 use part 1 of the fundamental theorem of calculus to find the derivative of the function.

1 answer:

7 0

$\displaystyle h(x)=\int\limits_{1}^{\sqrt{x}}~\cfrac{z^2}{z^4+1}dz~\hspace{10em}\cfrac{dh}{du}\cdot \stackrel{chain~rule}{\cfrac{du}{dx}\implies \cfrac{dh}{dx}} \\\\[-0.35em] ~\dotfill\\\\ u=\sqrt{x}\implies \cfrac{du}{dx}=\cfrac{1}{2\sqrt{x}} \\\\[-0.35em] ~\dotfill$

$\cfrac{dh}{dx}\implies \displaystyle \cfrac{d}{du}\left[ \int\limits_{1}^{u}~\cfrac{z^2}{z^4+1}dz \right]\cdot \cfrac{1}{2\sqrt{x}}\implies \left[ \cfrac{u^2}{u^4+1} \right]\cdot \cfrac{1}{2\sqrt{x}} \\\\\\ \stackrel{substituting~back}{\left[ \cfrac{(\sqrt{x})^2}{(\sqrt{x})^4+1} \right]\cdot \cfrac{1}{2\sqrt{x}}}\implies \cfrac{x}{x^2+1}\cdot \cfrac{1}{2\sqrt{x}}\implies \cfrac{\sqrt{x}}{2x^2+2}$

You might be interested in

Mona wants to buy carpet for a floor in her house.

Gnesinka [82]

What are the prices?

7 0

3 years ago

A string passing over a smooth pulley carries a stone at one end. While its other end is attached to a vibrating tuning fork and

nasty-shy [4]

Answer:

correct option is C) 2.8

Step-by-step explanation:

given data

string vibrates form = 8 loops

in water loop formed = 10 loops

solution

we consider mass of stone = m

string length = l

frequency of tuning = f

volume = v

density of stone = $\rho$

case (1)

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

so here

$l = \frac{8 \lambda _1}{2}$ ..............1

$l = 4 \lambda_1\\\\\lambda_1 = \frac{l}{4}$

and we know velocity is express as

velocity = frequency × wavelength .....................2

$\sqrt{\frac{Tension}{mass\ per\ unit \length }}$ = f × $\lambda_1$

here tension = mg

$\sqrt{\frac{mg}{\mu}}$ = f × $\lambda_1$ ..........................3

and

case (2)

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

$l = \frac{10 \lambda _1}{2}$ ..............4

$l = 5 \lambda_1\\\\\lambda_1 = \frac{l}{5}$

when block is immersed

equilibrium eq will be

Tenion + force of buoyancy = mg

T + v × $\rho$ × g = mg

and

T = v × $\rho$ - v × $\rho$ × g

from equation 2

f × $\lambda_2$ = f × $\frac{1}{5}$

$\sqrt{\frac{v\rho _{stone} g - v\rho _{water} g}{\mu}} = f \times \frac{1}{5}$ .......................5

now we divide eq 5 by the eq 3

$\sqrt{\frac{vg (\rho _{stone} - \rho _{water})}{\mu vg \times \rho _{stone}}} = \frac{fl}{5} \times \frac{4}{fl}$

solve irt we get

$1 - \frac{\rho _{stone}}{\rho _{water}} = \frac{16}{25}$

relative density $\frac{\rho _{stone}}{\rho _{water}} = \frac{25}{9}$

relative density = 2.78 ≈ 2.8

so correct option is C) 2.8

3 0

3 years ago

At the grocery, peaches cost $2.60 per pound. Grace bought 2.3 pounds of peaches.

brilliants [131]

The total cost would be $5.98
hope this helps! :)

3 0

3 years ago

1.) You are one of five people who have been hired to mow the lawn at the local college. Each week, the dimensions of the plot y

iren [92.7K]

Answer:

1a) Length = 7x + 3 & Width = 4x - 2

1b) Area = $28x^2 -2x -6$

1c) Area = 2774 sq. m

2. $(x-4)(y+7)$

Step-by-step explanation:

1a)

The length given as words is "3 more than 7 times x"

The width given as words is "4 times x minus 2"

The expression for length would be 7x + 3

The expression for width would be 4x - 2

1b)

The area is length * width

Since we already know the algebraic expressions for length and width from part (a) above, we use the formula:

Area = (7x+3)(4x-2) = 28x^2 -14x + 12x - 6 = 28x^2 -2x -6

Area = $28x^2 -2x -6$

1c)

Given x = 10, we put this into the area expression we found in (b) above.Let's see:

$28x^2 -2x -6\\=28(10)^2 -2(10) -6\\=2774$

Area = 2774 sq. m

We can group the first two terms and next two terms and write up:

$xy-4y+7x-28\\(xy-4y)+(7x-28)\\y(x-4)+7(x-4)\\(x-4)(y+7)$

That's the factored form.

8 0

3 years ago

Read 2 more answers

If the numerator of a fraction is increased by 3, the fraction becomes 3/4. If the denominator is decreased by 7, the

Sauron [17]

7 over 1

Step-by-step explanation:

3 0

3 years ago