The value of k is
<h3>How to solve the simultaneous equation?</h3>
Given:
x-y=k.............(eq i)
2x²+y²-15..............(eq ii)
We would make y the subject formula in eq ii
2x²+y²-15= 0
2x² + y²= 15
y²= 15-2x²
y= 
...........(eq iii)
Substitute the value of y into eq i
x-(= k
x- (
= k
k=
Read more about simultaneous equations here:
brainly.com/question/16863577
#SPJ1
35-4x>2
-4x>2-35
-4x>-33
-x>-33/4
x<33/4 (x<8.25)
Solution: x<8.25. or x∈(-∞ , 8.25)
416.25
375*.11=41.25
41.25+375=416.25
X=6 and x=1
You can find your zeros by determining what you have to plug into the function in order for it to equal zero
If we plug in 6, for example we’d get (6-6)(x-1)
Simplified this is 0(x-1)
Anything times 0 is 0, so this is one of our zeros.
Same goes for x-1, we just need to plug in 1 for it to equal 0
Therefore there are zeros at x=1 and x=6 :))
Answer:
d. reflect triangle abc across line ab
Step-by-step explanation:
