<span>Multiply one of the equations so that both equations share a common complementary coefficient.
In order to solve using the elimination method, you need to have a matching coefficient that will cancel out a variable when you add the equations together. For the 2 equations given, you have a huge number of choices. I'll just mention a few of them.
You can multiply the 1st equation by -2/5 to allow cancelling the a term.
You can multiply the 1st equation by 5/3 to allow cancelling the b term.
You can multiply the 2nd equation by -2.5 to allow cancelling the a term.
You can multiply the 2nd equation by 3/5 to allow cancelling the b term.
You can even multiply both equations.
For instance, multiply the 1st equation by 5 and the second by 3. And in fact, let's do that.
5a + 3b = –9
2a – 5b = –16
5*(5a + 3b = -9) = 25a + 15b = -45
3*(2a - 5b = -16) = 6a - 15b = -48
Then add the equations
25a + 15b = -45
6a - 15b = -48
=
31a = -93
a = -3
And then plug in the discovered value of a into one of the original equations and solve for b.</span>
26 maybe it can be ım not sure
Answer:
you cant round that to 30
Step-by-step explanation
29.20
29.2 2 is less than 5
so you cant round you can only round if the .tens place is 5 or more
Answer:
Therefore the answer is 20.
Step-by-step explanation:
We know that
class interval = range / number of classes
But here number of classes is not given , so we use the formula
class interval = range / ( 1+ 3.322 log N)
where , range =maximum - minimum = 220-100 = 120
N= number of observations = 50
class interval = 120 / ( 1+ 3.322 * log 50) = 18.06
Rounding up to a convinient number
Thus , class intervai = 20
Therefore the answer is 20.
Answer:
A=45m^2
Step-by-step explanation:
l*w
