ΔAOB is a right angled triangle. Therefore the Pythagorean Theorem applies in this situation.
θ is the angle from a standard position of the line OA
The length of the y component is √(1-0)2 +(-3-(-3))2] =√(12+ 02) = 1 A(-3,1) to B(-3,0) which is opposite
Then the length of the x-component is √[(-3-0)2 +(0-0)2] = √(9+0)= 3 B(-3,0) to O(0,0) which is adjacent
The length of vector OA is √[(-3-0)2 + (1-0)2] = √(9+1) = √(10) A(-3,1) to O(0,0) which is the hypotenuse of the triangle
θ = 180 - α
sinθ = sin(180-α) = opposite/hypotenuse = 1/√10
cosθ = adjacent/hypotenuse = -3/√10
tanθ = opposite/adjacent = 1/-3 = -1/3
α= arcsin(1/√10) ≈ 18
θ =180 -18 ≈162
Answer:
y = -x - 2
Step-by-step explanation:
y = mx + c
m = (-1-3)/(-1-(-5)) = -4/4 = -1
y = -x + c
(-1,-1)
-1 = -(-1) + c
c = -2
y = -x - 2
Answer:
A perfect square is a whole number that is the square of another whole number.
n*n = N
where n and N are whole numbers.
Now, "a perfect square ends with the same two digits".
This can be really trivial.
For example, if we take the number 10, and we square it, we will have:
10*10 = 100
The last two digits of 100 are zeros, so it ends with the same two digits.
Now, if now we take:
100*100 = 10,000
10,000 is also a perfect square, and the two last digits are zeros again.
So we can see a pattern here, we can go forever with this:
1,000^2 = 1,000,000
10,000^2 = 100,000,000
etc...
So we can find infinite perfect squares that end with the same two digits.
The answer is C multiply 3x2 and multiply 4x3 then add hope that helps
