Mathematics, 04.01.2020 04:31 ReaganS4
Find the value of z such that the line x=z divides the region bounded by the graphs of the equations into 2 regions of equal area.
y²=4-x, x=0 is one bound
the right bound is x=2
show all work and reasoning why you did it
use integrals
Answers: 3
![A_1=A_2\\\\ 2\displaystyle\int^z_0\sqrt{4-x}\,dx}=2\displaystyle\int^2_z\sqrt{4-x}\,dx}\\\\ \displaystyle\int^z_0\sqrt{4-x}\,dx}=\displaystyle\int^2_z\sqrt{4-x}\,dx}\\\\ \left[-\dfrac{2}{3}(4-x)^{\frac{3}{2}}\right]^z_0=\left[-\dfrac{2}{3}(4-x)^{\frac{3}{2}}\right]^2_z\\\\ \left[(4-x)^{\frac{3}{2}}\right]^z_0=\left[(4-x)^{\frac{3}{2}}\right]^2_z\\\\](/tpl/images/0442/1547/f7ce7.png)
![(4-z)^{\frac{3}{2}}-(4-0)^{\frac{3}{2}}=(4-2)^{\frac{3}{2}}-(4-z)^{\frac{3}{2}}\\\\ (4-z)^{\frac{3}{2}}-4^{\frac{3}{2}}=2^{\frac{3}{2}}-(4-z)^{\frac{3}{2}}\\\\ 2(4-z)^{\frac{3}{2}}=4^{\frac{3}{2}}+2^{\frac{3}{2}}\\\\ 2(4-z)^{\frac{3}{2}}=8+2\sqrt2\\\\ (4-z)^{\frac{3}{2}}=4+\sqrt2\\\\ ((4-z)^{\frac{3}{2}})^2=(4+\sqrt2)^2\\\\ (4-z)^3=16+8\sqrt2+2\\\\ (4-z)^3=18+8\sqrt2\\\\ 4-z=\sqrt[3]{18+8\sqrt2}\\\\ \boxed{z=4-\sqrt[3]{18+8\sqrt2}}](/tpl/images/0442/1547/24c77.png)
![[will give brainliest] given o below, is pq a minor arc, a major arc, or a semicircle?](/tpl/images/02/06/nZ4DHjbFPBA3siW1.jpg)
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Find the value of z such that the line x=z divides the region bounded by the graphs of the equations...
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