Answer:
Should be 2, -5
Step-by-step explanation:
you flip –2, 5 along y axis, then get 2, 5
then flip 2, 5 along x axis and get 2, -5
feel free to correct
Answer:
The sum of any integer and its opposite is equal to zero. adding two negative integers always yields a negative sum. To find the sum of a positive and a negative integer, take the absolute value of each integer and then subtract these values.
Answer:
(2, 1)
Step-by-step explanation:
The best way to do this to avoid tedious fractions is to use the addition method (sometimes called the elimination method). We will work to eliminate one of the variables. Since the y values are smaller, let's work to get rid of those. That means we have to have a positive and a negative of the same number so they cancel each other out. We have a 2y and a 3y. The LCM of those numbers is 6, so we will multiply the first equation by a 3 and the second one by a 2. BUT they have to cancel out, so one of those multipliers will have to be negative. I made the 2 negative. Multiplying in the 3 and the -2:
3(-9x + 2y = -16)--> -27x + 6y = -48
-2(19x + 3y = 41)--> -38x - 6y = -82
Now you can see that the 6y and the -6y cancel each other out, leaving us to do the addition of what's left:
-65x = -130 so
x = 2
Now we will go back to either one of the original equations and sub in a 2 for x to solve for y:
19(2) + 3y = 41 so
38 + 3y = 41 and
3y = 3. Therefore,
y = 1
The solution set then is (2, 1)
Answer:
We have to prove
sin(α+β)-sin(α-β)=2 cos α sin β
We will take the left hand side to prove it equal to right hand side
So,
=sin(α+β)-sin(α-β) Eqn 1
We will use the following identities:
sin(α+β)=sin α cos β+cos α sin β
and
sin(α-β)=sin α cos β-cos α sin β
Putting the identities in eqn 1
=sin(α+β)-sin(α-β)
=[ sin α cos β+cos α sin β ]-[sin α cos β-cos α sin β ]
=sin α cos β+cos α sin β- sinα cos β+cos α sin β
sinα cosβ will be cancelled.
=cos α sin β+ cos α sin β
=2 cos α sin β
Hence,
sin(α+β)-sin(α-β)=2 cos α sin β
Answer:
- 670 adult tickets
- 1340 student tickets
Step-by-step explanation:
One group of 2 student and 1 adult tickets will go for 2×$12 +16 = $40. The number of such groups sold was ...
$26,800/$40 = 670
There were 670 adult tickets sold.
There were 670×2 = 1340 student tickets sold.
