<h3>E
xplanation:</h3>
Replace cos^2(θ) with 1-sin^2(θ), and cot(θ) with cos(θ)/sin(θ).
cos^2(θ)cot^2(θ) = cot^2(θ) - cos^2(θ)
(1 -sin^2(θ))cot^2(θ) = . . . . . replace cos^2 with 1-sin^2
cot^2(θ) -sin^2(θ)·cos^2(θ)/sin^2(θ) = . . . . . replace cot with cos/sin
cot^2(θ) -cos^2(θ) = cot^2(θ) -cos^2(θ) . . . as desired
Answer:
D: 7
Step-by-step explanation:
1.75 (7) + 4.50
12.25 + 4.50
= 16.75
16.75 < 20
Answer:
The sum is a binomial with a degree of 6
Step-by-step explanation:
we have
Group terms that contain the same variable
The sum is a binomial ( two terms) with a degree of 6
has a degree of 6 (x has an exponent of 1, y has 5, and 1+5=6)
Answer:
Step-by-step explanation:
5x+20 = 0
5X= -20
X= -20/5 = -4
8(4n+3) = 32n+24 =0
32n = -24
n = -24/32 = -3/4
12n-6=0
12n =6
n = 6/12 = 1/2
9(2x+9)
18x+81 =0
18x= -81
x= -81/18 = -9/2= -4.5
36+9y=0
36=-9y
36/-9 = y = -4
7(n-3) = 0
7n-21=0
7n = 21
n = 21/7
n=3
21+15a = 0
21= -15a
a= 21/-15 = 7/-5 = -1.4
32x-12x=0
20x=0
x=0/20 = 0
n-4-9=0
n-13=0
n=13
-3x-9+15x=0
-3x+15x-9 =0
12x-9=0
12x=9
x=9/12 = 3/4 = 0.75
-16N-14N=0
-30n=0
n=0/-30 =0
-10-90x-6x-10x²-2
-10-2-96x-10x²
-12-96x-10x²
10x²+96x+12
x= -b±√b²-4(a)(c)/2a
{-96±√96²-4(10)(12)}/2(10)
{-96±√(9216-480)}/20
-96±√8736/20
-96±(93.4665715644)/20
-96+93.4665715644/20 -96-93.4665715644/20
-2.5334284356/20 -189.466571564/20
-0.12667142178≅ -0.13 -9.4733285782≅ -9.47
-0.13 -9.47
-5(9n+9) =0
-45n-45 =0
-45n = 45
n = 45/-45 =-1
-5n+3(6+7n)
-5n+18+21n=0
-5n+21n+18 =0
16n+18 =0
16n = -18
n = -18/16 = -9/8 = -1.125
-4+7(1-3n)=0
-4+7-21n=0
3-21n=0
3=21n
3/21=n = 1/7
3n-(2n-8)
3n-2n+8=0
n+8=0
n=-8
