$\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)$

$(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166$

Hence, we have that the amount dumped after 5 days is 0.166 tons.

Have a nice day!

5 0

4 years ago

Find the coordinates of the points of intersection of the graph of y=13−x with the axes and compute the area of the right triang

Ymorist [56]

the coordinates of the points of intersection of the graph of y=13−x with the axes

Given equation is y=13-x

x axis is x=0

y axis is y=0

Plug in the value of x =0 and find out y

y = 13 - x

y = 13 - 0 = 13

Plug in the value of y=0 and find out x

0 = 13 - x

x= 13

So coordinate axis

x= (13,0)

y= (0,13)

Base of the triangle is x=13

Height of the triangle is y= 13

Area of the triangle = $\frac{1}{2} * base * height$

= $\frac{1}{2} * 13 * 13$ = 84.5

Area of the triangle = 84.5

6 0

3 years ago

Solve for v. 9 2 V-2 > 6​

Solve for v. 9 2 V-2 > 6