Answer:
d
Step-by-step explanation:
Some are
y-2=-3x
y+3x=2
3x=2-y
Answer: 2,944,321
Step-by-step explanation: Use long addition and evaluate.
Hope this helps you out! ☺
-Leif Jonsi-
Step-by-step explanation:

In this method , the second order of polynomial ax² + bx + c is factorised and expressed as the product of two linear factors. Then each linear factor is separately solved to get the required solutions of the equation by applying zero factor property. In zero factor property , if p•q = 0 , then either p = 0 or q = 0 . In other words , if the product of two numbers is 0 , then one or both of the numbers must be 0.


Here , we have to find the two numbers that multiplies to 55 and adds to 16.
⤑ 
⤑ 
⤑ 
⤑ 
Either :



Hope I helped ! ♡
Have a wonderful day / night ! ツ
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You cannot assume the angles add to 90, but you know since BD is an angle bisector that ABD is equal to DBC, or x-5=2x-6
x=1. this is the correct solution to the equation but gives negative angles when plugged in which isn't possible. there must be something wrong work the question