As hamster wheel is circular so for revolution we will find the circumference.
Diameter of the Hamster wheel = 8.5 in
One revolution of the wheel = circumference of the wheel
Circumference = π × d
π = 3.14 and d = 8.5
So, Circumference = 3.14 × 8.5
Circumference = 26.69 in
So in one revolution the hamster wheel travels a distance of 26.69 in
Answer:
Step-by-step explanation:
hello :
-4+5|4x-4|=21
add 4 : 5|4x-4| = 25
divid by 5 : |4x-4| = 5 means : 4x-4 = 5 or 4x-4 = - 5
4x=9 or 4x =-1
x= 9/4 or x = - 1/4
I used to hate fractions. But in time, you learn to love them. This is because there's a big difference between fractions and decimals, even though when you divide the actual fraction it comes out to a decimal. Decimals go on and on sometimes, and it would be impossible to write out all those numbers, especially when taking a timed test, for example. Fractions, in this case, would be much more useful (as long as you know how to use them to your advantage). Fractions are basically all those decimal numbers wrapped up into a single, simple division. It makes the outcome of your answer much more accurate than if you estimate every decimal you get throughout a math problem. The more you estimate throughout the problem-solving process, the less accurate your final answer will be. Hence why teachers will usually tell you to estimate when you're putting down the final answer. Fractions are complex at times, so it may be easier to use them in decimal form for certain situations (especially if the decimal form is short and sweet). A world without fractions will result in many, many inaccurate situations involving mathematical knowledge.
Answer:
espero que esto ayude
Step-by-step explanation:
length = int (input ("Ingresa una longitud"))
width = int (input ("Ingresa un ancho"))
área = largo * ancho
imprimir ("El área es" + área)
The acceleration of the particle is given by the formula mentioned below:
Differentiate the position vector with respect to t.
![\begin{gathered} \frac{ds(t)}{dt}=\frac{d}{dt}\sqrt[]{\mleft(t^3+1\mright)} \\ =-\frac{1}{2}(t^3+1)^{-\frac{1}{2}}\times3t^2 \\ =\frac{3}{2}\frac{t^2}{\sqrt{(t^3+1)}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cfrac%7Bds%28t%29%7D%7Bdt%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Csqrt%5B%5D%7B%5Cmleft%28t%5E3%2B1%5Cmright%29%7D%20%5C%5C%20%3D-%5Cfrac%7B1%7D%7B2%7D%28t%5E3%2B1%29%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D%5Ctimes3t%5E2%20%5C%5C%20%3D%5Cfrac%7B3%7D%7B2%7D%5Cfrac%7Bt%5E2%7D%7B%5Csqrt%7B%28t%5E3%2B1%29%7D%7D%20%5Cend%7Bgathered%7D)
Differentiate both sides of the obtained equation with respect to t.
![\begin{gathered} \frac{d^2s(t)}{dx^2}=\frac{3}{2}(\frac{2t}{\sqrt[]{(t^3+1)}}+t^2(-\frac{3}{2})\times\frac{1}{(t^3+1)^{\frac{3}{2}}}) \\ =\frac{3t}{\sqrt[]{(t^3+1)}}-\frac{9}{4}\frac{t^2}{(t^3+1)^{\frac{3}{2}}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cfrac%7Bd%5E2s%28t%29%7D%7Bdx%5E2%7D%3D%5Cfrac%7B3%7D%7B2%7D%28%5Cfrac%7B2t%7D%7B%5Csqrt%5B%5D%7B%28t%5E3%2B1%29%7D%7D%2Bt%5E2%28-%5Cfrac%7B3%7D%7B2%7D%29%5Ctimes%5Cfrac%7B1%7D%7B%28t%5E3%2B1%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%29%20%5C%5C%20%3D%5Cfrac%7B3t%7D%7B%5Csqrt%5B%5D%7B%28t%5E3%2B1%29%7D%7D-%5Cfrac%7B9%7D%7B4%7D%5Cfrac%7Bt%5E2%7D%7B%28t%5E3%2B1%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%20%5Cend%7Bgathered%7D)
Substitute t=2 in the above equation to obtain the acceleration of the particle at 2 seconds.
![\begin{gathered} a(t=1)=\frac{3}{\sqrt[]{2}}-\frac{9}{4\times2^{\frac{3}{2}}} \\ =1.32ft/sec^2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20a%28t%3D1%29%3D%5Cfrac%7B3%7D%7B%5Csqrt%5B%5D%7B2%7D%7D-%5Cfrac%7B9%7D%7B4%5Ctimes2%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%20%5C%5C%20%3D1.32ft%2Fsec%5E2%20%5Cend%7Bgathered%7D)
The initial position is obtained at t=0. Substitute t=0 in the given position function.
