4 sin^2 θ + 13cos^2 θ = 7
sin^2 θ = 1 - cos^ θ
4 - 4cos^2 θ + 13cos^2 θ = 7
9cos^2 θ = 3
cos^2 θ = 1/3
cos θ = # (1/3) # - square root
Square root of (1/3) has +1/3 and -1/3 as values of cos
Find the key angle by doing the cos inverse of #1/3
K.A = cos^-1 #(1/3) = 0.955
θ lies in all 4 quadrants
The values of θ are:
θ = 0.955, 2.186, 4.096, 7.23
Ignore 0.955, 2.186, 4.096, 7.23 as they are out of range pi/2 = 1.571
The the value of θ = 0.955 = 0.96 (to 2 d.p) radian
Hope it helped!
Hello! Let's work this problem from left to right to make things a bit easier ;)
6.5+(-2) is the same as 6.5-2, since adding a negative is the same as subtracting a positive. 6.5-2=4.5
Now, we have to add 10.5 to 4.5, which gives us the grand total of 15. I hope this helped! Let me know if you need anything else!
First, for end behavior, the highest power of x is x^3 and it is positive. So towards infinity, the graph will be positive, and towards negative infinity the graph will be negative (because this is a cubic graph)
To find the zeros, you set the equation equal to 0 and solve for x
x^3+2x^2-8x=0
x(x^2+2x-8)=0
x(x+4)(x-2)=0
x=0 x=-4 x=2
So the zeros are at 0, -4, and 2. Therefore, you can plot the points (0,0), (-4,0) and (2,0)
And we can plug values into the original that are between each of the zeros to see which intervals are positive or negative.
Plugging in a -5 gets us -35
-1 gets us 9
1 gets us -5
3 gets us 21
So now you know end behavior, zeroes, and signs of intervals
Hope this helps<span />
Polynomial Degree:
3
3
Leading Term:
5
n
3
5
n
3
Leading Coefficient:
5
Answer:
Jessiah has 14 pencils
Miles has 28 pencils
Grace has 8 pencils
Step-by-step explanation:
I'm going to assume that you mean Jessaih instead of Jessia because otherwise it would be impossible to solve
Let the # of Jessaih's pencils equal j, # of Grace's pencils equal g, # of Miles' pencils equal m,
j=g+6
m=2j
So m=2(g+6) or 2g+12
Therefore, g+g+6+2g+12, or 4g+18=50, so g = 8
j=8+6, which is 14
m = 2(14), which is 28
