Answer:
Answers: 1) gh= -7×3= -21
2) g2-h= -7×2-3= -17 *that is what I got from my calculations.
3) g+ h2 = -7+3×2= 1 *that is what I got from my calculations.
4) g+h = -7+3 = -4
5) h-g= 3--7= 3+7= 10
6) g-h= -7-3= 10
Step-by-step explanation:
I don't fully understand neither but I tried.
This is not an identity.
Check x = π/4, for which we have cos(π/4) = sin(π/4) = 1/√2. Together with sin(2•π/4) = sin(π/2) = 1 and cos(2•π/4) = cos(π/2) = 0, the left side becomes 1, while sec(π/4) = 1/cos(π/4) = √2.
Keeping the left side unchanged, the correct identity would be
To show this, recall
• sin(2x) = 2 sin(x) cos(x)
• cos(2x) = cos²(x) - sin²(x)
• cos²(x) + sin²(x) = 1
Then we have
Extranious solution is mostly when you take the square root of negative number
what you want to do is math stuff
translate
remember that
so
7.
add 2 to both sides
square both sides
cube root both sides
x+1=9
minus 1
x=8
8.
square both sides
using math
0=x²-5x-6
x=6 or -1
if we do x=-1, we get
√4=-2, which is kind of true, but we normally find the principal root (posiitive0
-1 is extraious
x=6
9.
add √(x-1) to both sides then square both sides
minus (x+5) from both sides
square both sides
using math
x=-1 or 15
if we had x=-1, we would be having √(-1-1) which is sqrt of negative which is not allowed
x=15
2 times 1/6 = 1/3
2/1 x 1/6 = 2/6, reduce this equals 1/3
The arrival time in london of a train that leaves Leicester to london at 11:50 A.M is 13:05 pm.
To calculate the arrival time in london of a train that leaves Leicester to london at 11:50 A.M, use use the formula below.
Formula:
- Ta = [(d/s)+11:50]....... Equation 1
Where:
- Ta = Arrival time
- d = distance from Leicester to London
- s = Speed
From the question,
Given:
Substitute these values into equation 1
- Ta = [(100/80)+11:50]
- Ta = 1.25h+11:50
- Ta = 1 hours 15 minutes +11:50
- Ta = 13:05 pm
Hence, the arrival time in london of a train that leaves Leicester to london at 11:50 A.M is 13:05 pm.
Learn more about time here: brainly.com/question/4931057