Answer:
Step-by-step explanation:
Use the half angle identity for cosine:
cos(x/2)=+ or - sqrt(1+cos(x))/sqrt(2)
I'm going to figure out the sign part first for cos(x/2)...
so x is in third quadrant which puts x between 180 and 270
if we half x, x/2 this puts us between 90 and 135 (that's the second quadrant)
cosine is negative in the second quadrant
so we know that
cos(x/2)=-sqrt(1+cos(x))/sqrt(2)
Now we need cos(x)... since we are in the third quadrant cos(x) is negative...
If you draw a reference triangle sin(x)=3/5 you should see that cos(x)=4/5 ... but again cos(x)=-4/5 since we are in the third quadrant.
So let's plug it in:
cos(x/2)=-sqrt(1+4/5)/sqrt(2)
No one likes compound fractions (mini-fractions inside bigger fractions)
Multiply top and bottom inside the square roots by 5.
cos(x/2)=-sqrt(5+4)/sqrt(10)
cos(x/2)=-sqrt(9)/sqrt(10)
cos(x/2)=-3/sqrt(10)
Rationalize the denominator
cos(x/2)=-3sqrt(10)/10
9514 1404 393
Answer:
(d) 8 cm and 9 cm
Step-by-step explanation:
The sum of the other two side lengths must <em>exceed</em> 13 cm for a triangle to be possible.
8 cm and 9 cm
This means Square root (^1/2). this is kind of hard to read. anyways. 32^(1/2) separates into 4^(1/2) and 8^(1/2), which further separates into 4^(1/2) and 2^(1/2), root 4 becomes 2, their are two root 4's, so you get 2 x 2, and you are left with 2^(1/2) now why go to all this trouble. because now you can multiply the 4 you created (2x2) times, the 7... giving you 28*2^(1/2) now subtract it from the the other one with root 2. -5*2^(1/2), giving you 23*2^(1/2)-...idk what that last bit is. if its a odd number then this is the end of the problem, if you can get it to root 2. then do that and simplify.
A. Let us first assume that the gas acts like an ideal
gas so that we can use the ideal gas equation:
PV = nRT
where P is pressure, V is volume, n is number of moles, R
is universal gas constant and T is absolute temperature
In this case, let us assume that nRT = constant so that:
P1V1 = P2V2
400 mm Hg * 400 mL = P2 * 200 mL
P2 = 800 mm Hg
<span>B. The collision of gas with the walls of the container
produces Pressure.</span>
