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3 years ago

15

What is

} of \frac{1}{4}" alt="\frac{1}{2} of \frac{1}{4}" align="absmiddle" class="latex-formula">

2 answers:

lidiya [134]3 years ago

8 0

Therefore, half of 1/4 is 1/8. 1/4 means 0.25 if you devide it by 2 you will get 0.125 so 1/8 is half of 1/4.

Sedbober [7]3 years ago

4 0

1/2 of 1/4 is 1/8. Explanation: 1/2 of 1/4 is the same thing as 1/2(1/4). 1/2*1/4 is 1/8.

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sharon serves the volleyball to rhonda with an upward velocity of 18.5 ft/s. The ball is 4.5 feet above the ground when she stri

andreyandreev [35.5K]

we can use height formula

$h(t)=-\frac{1}{2} g t^2+vt+h(0)$

where

h(t) is the height after t seconds

v is velocity

h(0) is initial height

g is acceleration due to gravity

we know that

g=32 ft/s^2

we have

v=18.5ft/s

h(0)=4.5ft

so, we can plug these values

and we get

$h(t)=-\frac{1}{2}*32 t^2+18.5t+4.5$

$h(t)=-16 t^2+18.5t+4.5$

now, we have to find time when object will hit the ground

and we know that object will hit the ground when height becomes 0

so, we can set h(t)=0

and then we can solve for t

$0=-16 t^2+18.5t+4.5$

we can use pythagoras theorem

$t=\frac{-185-\sqrt{185^2-4\left(-160\right)45}}{2\left(-160\right)}=\frac{37+\sqrt{2521}}{64}$

$t=1.36265sec$ ..............Answer

7 0

3 years ago

Eli is driving on a long road trip. He currently has 12 gallons of gas in his car. Each

AlladinOne [14]

Answer:

I would be 7 gallons used and 5 gallons left

Step-by-step explanation:

multyply 1.75 times 4 hours equal 7 then 12-7=5 is the total of gallons left

Hope this helps

5 0

3 years ago

Read 2 more answers

Elsies math class has 11 female students and 10 male students. Write the ratio of female students to total number of students in

stiv31 [10]

Answer:

Step-by-step explanation:

7 0

3 years ago

Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

7 0

3 years ago

11×11=4 22×22=16 33×33=?

Kay [80]

This is your answer <span>33 x 33 = 4^3 = 64</span>

7 0

3 years ago