Answer:
we have
secθ = 2
=2
Cosθ =
b=1
h=2
p=
again
Cot θ=
Cot θ=
Cot θ=
It lies in Quadrant III cot is positive
Cot θ=
<u>b. (sqrt 3)/3</u>
( 4 + 2 , - 5 - 8 ) = ( 6 , - 13 )
The last option ...
Prove sin^6 a + cos^6 a =1-3sin^2 a cos^ a
1) sin^6 a + cos^6 a = (sin^2 a)^3 + (cos^2 a)^3
2) a^3 + b^3 = (a+b) (a^2 - ab + b^2)
3) (sin^2 a)^3 + (cos^2 a)^3 = (sin^2 a + cos^2 a) (sin^4 a - sin^2 a* cos^2 a) + cos<span>^4 a)
4) </span><span>(sin^2 a + cos^2 a) = 1
5 ) </span>sin^6 a + cos^6 a = sin^4 a + cos^4 a - sin^2 a* cos^2 a<span>
6) a</span>^4 + b^4 = (a^2 +b^2 )- 2 a^2 b^2
7) sin^4 a + cos^4 a -sin^2 a* cos^2 a =(sin^2 a + cos^2 a)-2sin^2 acos<span>^2 a
8) </span>sin^6 a + cos^6 a = 1- 2sin^2 a cos^2 a - <span>sin^2 a* cos^2 a
9) </span><span>sin^6 a + cos^6 a = 1- 3 sin^2 a cos^2 a</span>
Answer:
The two numbers are -5 and -4
Explanation:
Assume that the first number is x and that the second number is x+1.
We know that the sum of their squares is 41. This means that:
x² + (x+1)² = 41
We will expand the brackets and factorize to get the value of x as follows:
x² + (x+1)² = 41
x² + x² + 2x + 1 = 41
2x² + 2x + 1 - 41 = 0
2x² + 2x - 40 = 0
We can divide all terms by 2 to simplify the equation:
x² + x - 20 = 0 ..........> equation required in part II
Now, we can factorize this equation to get the values of x:
x² + x - 20 = 0
(x-4)(x+5) = 0
either x = 4 .........> rejected because we know that x should be negative
or x = -5 ...........> accepted
Based on the above calculations, the two numbers are -5 and -4
Hope this helps :)
Answer:
Step-by-step explanation:
Use slope formula.
Rise over run.
Or
y over x.
Over changes in x value would be - 3/4 pi.
Plug in seepage intervals for x to find y.
In the regular function,
Since our period is 2, it would stay the same since 1x2=2
Since our amplitude is 6, our y value now is 6.
Since our vertical shift is -4, our y value is 2.
So
Let do the other point,
Our period is 2 so
Multiply this by 6.
It stays 0 then subtract 4 we get
Use the earlier formula, slope