Answer:
y = 60
Step-by-step explanation:
Given that y varies directly with x then the equation relating them is
y = kx ← k is the constant of variation
To find k use the condition y = 20 when x = 10, then
20 = 10k ( divide both sides by 10 )
2 = k
y = 2x ← equation of variation
When x = 30, then
y = 2 × 30 = 60
Answer: y=12.287
Step-by-step explanation:
Answer:
1024/1771561
Step-by-step explanation:
Using exponent rules, we can multiply all of them together after applying the exponent to get the final answer.
Check the picture below.
![\stackrel{\textit{\Large Areas}}{\stackrel{triangle}{\cfrac{1}{2}(6)(6)}~~ + ~~\stackrel{semi-circle}{\cfrac{1}{2}\pi (3)^2}}\implies \boxed{18+4.5\pi} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{pythagorean~theorem}{CA^2 = AB^2 + BC^2\implies} CA=\sqrt{AB^2 + BC^2} \\\\\\ CA=\sqrt{6^2+6^2}\implies CA=\sqrt{6^2(1+1)}\implies CA=6\sqrt{2} \\\\\\ \stackrel{\textit{\Large Perimeters}}{\stackrel{triangle}{(6+6\sqrt{2})}~~ + ~~\stackrel{semi-circle}{\cfrac{1}{2}2\pi (3)}}\implies \boxed{6+6\sqrt{2}+3\pi}](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Ctextit%7B%5CLarge%20Areas%7D%7D%7B%5Cstackrel%7Btriangle%7D%7B%5Ccfrac%7B1%7D%7B2%7D%286%29%286%29%7D~~%20%2B%20~~%5Cstackrel%7Bsemi-circle%7D%7B%5Ccfrac%7B1%7D%7B2%7D%5Cpi%20%283%29%5E2%7D%7D%5Cimplies%20%5Cboxed%7B18%2B4.5%5Cpi%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7Bpythagorean~theorem%7D%7BCA%5E2%20%3D%20AB%5E2%20%2B%20BC%5E2%5Cimplies%7D%20CA%3D%5Csqrt%7BAB%5E2%20%2B%20BC%5E2%7D%20%5C%5C%5C%5C%5C%5C%20CA%3D%5Csqrt%7B6%5E2%2B6%5E2%7D%5Cimplies%20CA%3D%5Csqrt%7B6%5E2%281%2B1%29%7D%5Cimplies%20CA%3D6%5Csqrt%7B2%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7B%5CLarge%20Perimeters%7D%7D%7B%5Cstackrel%7Btriangle%7D%7B%286%2B6%5Csqrt%7B2%7D%29%7D~~%20%2B%20~~%5Cstackrel%7Bsemi-circle%7D%7B%5Ccfrac%7B1%7D%7B2%7D2%5Cpi%20%283%29%7D%7D%5Cimplies%20%5Cboxed%7B6%2B6%5Csqrt%7B2%7D%2B3%5Cpi%7D)
notice that for the perimeter we didn't include the segment BC, because the perimeter of a figure is simply the outer borders.
Answer:
The Acceleration of bike after 3 minutes of ride is 14 meter² per minutes
Step-by-step explanation:
Given as :
The displacement of bike is the function of time
i.e s(t) = t³ - 2 t² + 3 meters
The Time of bike acceleration = t = 3 minutes
Let The Acceleration of bike = a meter² per minutes
Now, According to question
∵ <u>Acceleration is define as the rate of change of velocity</u>
i.e acceleration = 
or, a = 
And
<u>Velocity is define as rate of change of displacement</u>
i.e velocity = 
Or, v = 
Since here displacement is function of time
So , v = 
Or, v = 
Or, v = 3 t² - 4 t + 3
So, velocity is the function of time = v = 3 t² - 4 t + 3 meter per minutes
Now, Again
Acceleration = a = 
Or, a = 
Or, a = 6 t - 4
∴ Acceleration is the function of time = a = 6 t - 4
Now, Acceleration of bike after 3 minutes
So, at t = 3 min
i.e a = 6 t - 4
Or, a = 6 × 3 - 4
Or, a = 18 - 4
∴ a = 14 meter² per minutes
Hence, The Acceleration of bike after 3 minutes of ride is 14 meter² per minutes . Answer