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I think the question asked about how the SYSTEM of the iss and capsule could evaluate the hypothesis, so i guess they wanted us to consider those two only rather than consider the Earth at all.

Imagine jacques's swim velocity as a vector of magnitude 1.5, and the current as a vector of magnitude 0.5. This question is basically asking what direction should jacques swim such that the two vectors add to become a vector perpendicular to the bank. something like this

for question 4 we are trying to find \int^{4}_{-4}f(x+3)dx as we know f(x+3) is basically f(x) shifted 3 units to the left, hence we can say that \int^{7}_{-1}f(x)dx=\int^{4}_{-4}f(x+3)dx or we can use the substitution u=x+3 to show the same thing now we can split \int^{7}_{-1}f(x)dx into...

Looks at the software, hardware, data, procedures and personnel of MYOB. I think it is alright.

the wavelength of light is \frac{3\times10^8}{5.36\times10^{14}}=559.70149...\times10^{-9} so basically the path difference is just the wavelength so there would be no change, hence B.

I heard they were making extra chapters for Solo Leveling a while back, but haven't checked at all. Also rip to Dubu, Solo Levelling was the first good manhuwa that I ever read, I will forever remember it. I see tessia is more your type.

oooh, so you were imagining sangah what happened to cha hae in?? 😬

This reminds me on Kim Dokja

Think of the pascal triangle and lets see if we can develop some patterns, as the pascal triangle can be developed through the combinations. as we can see, each of the numbers in the pascal triangle can be generated by adding the two numbers closest to it in the row above. ie 1+1=2, 1+2=3...

using the identity z\bar{z}=|z|^2 \begin{align*}\text{LHS}^2&=|z_1z_2+z_1z_3+z_2z_3|^2\\&=(z_1z_2+z_1z_3+z_2z_3)\overline{(z_1z_2+z_1z_3+z_2z_3)}\\&=|z_1|^2|z_2|^2+|z_1|^2z_2\bar{z_3}+z_1|z_2|^2\bar{z_2}+|z_1|^2\bar{z_2}z_3+|z_1|^2|z_3|^2+z_1\bar{z_2}|z_3|^2+\bar{z_1}|z_2|^2z_3+\bar...

yeah the roots would be \pm\cos\left(\frac{\pi}{12}\right),\pm\cos\left(\frac{5\pi}{12}\right),\pm\frac{1}{\sqrt{2}} lets say you start off with \cos\left(\frac{\pi}{12}\right) multiplying with \pm\cos\left(\frac{5\pi}{12}\right),\pm\frac{1}{\sqrt{2}} and then -\cos\left(\frac{\pi}{12}\right)...

use the sum to product of pairs of roots to show that \cos^2\left(\frac{\pi}{12}\right)+\cos^2\left(\frac{5\pi}{12}\right)=1 then use the fact that 2\cos\left(\frac{\pi}{12}\right)\cos^2\left(\frac{5\pi}{12}\right)=\frac{1}{2} to find...

let's just get this shit done with

the general equation of a circle is (x+a)^2+(y+b)^2=r^2 hence we will need at least 3 points to get an equation, we already have two. We can find another point by using the triangle ADC which is a right angle cause D is the midpoint of AB (x=-1) and angle CAD is \frac{\pi}{6} using trig we can...