What is the permiter problems first?? all you need to do to find perimeter is multiply the length by the width
Start by solving one equation for x or y, this time I'm choosing x. 2x = 4 - 2y, then x = 2 - y. Now substitute in x in the other equation. 4(2-y) - 3y = 15. This simplifies to 8 - 4y - 3y = 15. This simplifies to -7y = 7. Divide both sides by -7, and y = -1. Now plug the y value into the other equation: 2x + 2(-1) = 4. This simplifies to 2x = 6, and x = 3.
X = 3, Y = -1 or (3,-1)
Answer:
Number 1: x≈4.9412824
Number 2: x=3√10,−3√10x=310,-310
Decimal Form:
x=9.48683298...−9.48683298...
Number 3: x≈−5.04938819
Number 4: x=−1+2√6,−1−2√6x=-1+26,-1-26
Decimal Form:
x=3.89897948…,−5.89897948…
Number 5: x=−7+√93,−7−√93x=-7+93,-7-93
Decimal Form:
x=2.64365076…,−16.64365076…
Number 6: x=1+i√3910,1−i√3910
Hey there Shawn!
![\left[\begin{array}{ccc}\boxed{\boxed{ 5\frac{1}{3}= \frac{16}{3} }} \ as \ an \ improper \ fraction\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cboxed%7B%5Cboxed%7B%205%5Cfrac%7B1%7D%7B3%7D%3D%20%5Cfrac%7B16%7D%7B3%7D%20%20%7D%7D%20%5C%20as%20%5C%20an%20%5C%20improper%20%5C%20fraction%5Cend%7Barray%7D%5Cright%5D%20)
Hope this helps you budd!
