What i can see the total amount is wrong
Answer:
a triangle;
Three faces of the cube meet to form the vertex, so the cross section is a two-dimensional figure with three sides
Step-by-step explanation:
We have to evaluate cos(2θ), knowing that θ is in the first quadrant and in standard position P(u,v) = (3,4).
We can picture this as:
We can write the relation:
We now look at the identities to find cos(2θ):
There are many identities for cos(2θ), but this is expressed in the information we already know, so we can solve as:
Answer: cos(2θ) = -7/25
<u>2x + 3y = 1</u>
<u>y = 3x + 15</u>
There's not much you can do with the first equation, because it has
two variables in it ... 'x' and 'y' . No matter how much you move them
around, you'll never be able to get either one equal to just a number.
Is there any way you could get rid of one of the variables in the first
equation, and have just 1 letter in it to solve for ?
Absolutely ! The second equation tells you something that 'y' is <u>equal</u> to,
(3x + 15). "EQUAL" is very powerful. It means that wherever you see 'y',
you can put (3x + 15) in its place, and you won't change anything or
upset anything. One thing you can do is take that (3x + 15) from the <span>
2nd</span> equation, and put it right into the first equation in place of 'y'.
You'll see how that helps as soon as you do it.
First equation: <u>2x + 3y = 1</u>
Substitute for 'y' : 2x + 3(<em>3x + 15</em>) = 1
Remove parentheses: 2x + 3(3x) + 3(15) = 1
2x + 9x + 45 = 1
Combine the terms with 'x' in them: 11x + 45 = 1
Look what you have now ! An equation with only one variable in it !
Subtract 45 from each side: 11x = -44
Divide each side by 11 : <em> x = -4</em>
You're more than halfway there. Now you know what 'x' is,
and you can use it with either equation to find what 'y' is.
-- If you use it with the first equation: <u> 2x + 3y = 1</u>
Put in the value of 'x': 2(<em>-4</em>) + 3y = 1
Remove the parentheses: -8 + 3y = 1
Add 8 to each side: 3y = 9
Divide each side by 3 : <em> y = 3</em>
-- If you use it with the 2nd equation: <u>y = 3x + 15</u>
Put in the value of 'x' : y = 3(<em>-4</em>) + 15
Remove the parentheses: y = -12 + 15
Add numbers on the right side: <em> y = 3</em> (same as the other way)
So there's your solution for the system of two equations:
<em> x = -4</em>
<em> y = 3</em>
