Well just substitute y (the first equation) into the second equation.
So
5x-4(5-3x)=-3
Then just solve the equation from there.
5x-20+12x=-3
17x=17
x=1
and if you need y, y=2
Answer:
school b
Step-by-step explanation:
it has more
Answer:
Jayce should not use either option. Option 1 is likely to be based so that paperbacks are overrepresented, while Option 2 is likely to be biased so that e-books are overrepresented.
Step-by-step explanation:
<em>Given:</em>
<em>Jayce volunteers for the local library. The librarian wants to find out whether the patrons prefer paperback books or e-books. Jayce cannot decide which method to use for polling the patrons.</em>
<em>Option 1: Poll every fifth patron who enters the library on Paperback Lovers Day.</em>
<em>Option 2: Poll every third patron who enters the library on Technology Appreciation Day.</em>
<em />
<em>Since, Option 1: Poll every fifth patron who enters the library on Paperback Lovers Day. It biased because since it paperback lovers day thus, it would be likely that most people like the paperback which make it biased.</em>
<em>Since, Option 2: Poll every third patron who enters the library on Technology Appreciation Day. It biased because since it technology appreciation day thus, it would be likely that most people don't care about these stuff since people coming in are for technology appreciation.</em>
<em />
<em />
<em>Therefore, the answer is:</em>
Jayce should not use either option. Option 1 is likely to be based so that paperbacks are overrepresented, while Option 2 is likely to be biased so that e-books are overrepresented.
<u><em>Kavinsky</em></u>
Answer:
x = 2 ±i
Step-by-step explanation:
x^2 = 4x-5
Subtract 4x from each side
x^2 - 4x = 4x-5 -4x
x^2 - 4x= -5
Complete the square
Take the coefficient of the x term, divide by 2 and square it
-4/2 =2 2^2 =4
Add 4 to each side
x^2 -4x+4 = -5 +4
The left side is (x-coefficient of the x term/2)^2
(x-2)^2 = -1
Take the square root of each side
sqrt((x-2)^2) = sqrt(-1)
x-2 = ±i
Add 2 to each side
x-2+2 = 2 ±i
x = 2 ±i
