Answer:
26x³ - 12x² + 5x + 7
Step-by-step explanation:
- (2x³ - 3x + 11) - (3x² + 1) x (4 - 8x)
- 2x³ - 3x+11 - (12x² - 24x³ + 4 - 8x)
- 2x³ - 3x + 11 - 12x² + 24x³ - 4 + 8x
- 26x³ + 5x + 7 - 12x²
= 26x³ - 12x² + 5x + 7
Answer:
I believe your answer would be the following:
A) 250 miles in 5 hours.
C) 200 miles in 4 hours
E) 700 miles in 14 hour
Step-by-step explanation:
Unit rate = 50
A) 250 miles in 5 hours // 50*5 = 250 (CORRECT)
B) 300 miles in 5 hours // 50*5 ≠ 300 (WRONG)
C) 200 miles in 4 hours // 50*4 = 200 (CORRECT)
D) 240 miles in 4 hours // 50*4 ≠ 240 (WRONG)
E) 700 miles in 14 hours // 50*14 = 700 (CORRECT)
Hope this helps! :)
Answer:
length of diagonals are BD = 33.039 cm and AC = 38.84 cm
Step-by-step explanation:
given,
sides of parallelogram is 20 cm and 30 cm
Angle between them is 80°. opposite angle of parallelogram are same the opposite angle is same.
other angle = 360° - 80° - 80° + 2 x = 200
x = 100°
Using cosine law
c² = a² + b² - 2ab cos γ
In ΔDAB
BD² = 30² + 20 ² - 2 × 30 × 20 cos 80°
BD = 33.039 cm
now in ΔADC
c² = a² + b² - 2ab cos γ
AC² = 20² + 30² - 2 × 30 × 20 cos 100°
AC = 38.84 cm
length of diagonals are BD = 33.039 cm and AC = 38.84 cm
-9a + 11ad - 35a + ad = -44a + 12ad
hope this helps
So you can do this multiple ways, I'll do this the way that I think makes sense the l most easily.
Cos (0) = 1
Cos (pi/2)=0
Cos (pi) =-1
Cos (3pi/2)=0
Cos (2pi)=1
Now if you multiply the inside by 4, the graph oscillates more violently (goes up and down more in a shorter period).
But you can always reduce it.
Cos (0)= 1
Cos (4pi/2) = cos (2pi)=1
Cos (4pi) =Cos (2pi) =1 (Any multiple of 2pi ==1)
etc...
the pattern is that every half pi increase is now a full period as apposed to just a quarter of one. That's in theory.
Now that you know that, the identities of Cosine are another beast, but mathematically.
You have.
Cos (2×2t) = Cos^2 (2t)-Sin^2 (2t)
Sin^2 (t)=-Cos^2 (t)+1..... (all A^2+B^2=C^2)
Cos (2×2t) = Cos^2 (t)-(-Cos^2 (t)+1)
Cos (2×2t)= 2Cos^2 (2t) - 1
2Cos^2 (2t) -1= 2 (Cos^2(t)-Sin^2(t))^2 -1
(same thing as above but done twice because it's cos ^2 now)
convert sin^2
2Cos^2 (2t)-1 =2 (Cos^2 (t)+Cos^2 (t)-1)^2 -1
2 (2Cos^2(t)-1)^2 -1
2 (2Cos^2 (t)-1)(2Cos^2 (t)-1)-1
2 (4Cos^4 (t) - 2 (2Cos^2 (t))+1)-1
Distribute
8Cos^4 (t) -8Cos^2 (t) +1
Cos (4t) =8Cos^4-8Cos^2 (t)+-1
