Answer:
1.f(x)=2x-5
i will take the set {-2,-1,0,1,2}
f(-2)=2(-2)-5
=-4-5
=-9
f(-1)=2(-1)-5
=-2-5
=-7
f(0)=2(0)-5
=-5
f(1)=2(1)-5
=-3
f(2)=2(2)-5
=-1
so the coordinates of the function is {-9,-7,-5,-3,-1}
2.f(x)=-3x+6
i will the take the set {-2,-1,0,1,2} too
f(-2)=-3(-2)+6=6+6=12
f(-1)=-3(-1)+6=3+6=9
f(0)=-3(0)+6=6
f(1)=-3(1)+6=-3+6=3
f(2)=-3(2)+6=-6+6=0
{12,9,6,3,0}
3.f(x)=2/3.x+4
{-2,-1,0,1,2}
f(-2)=2/3(-2)+4=-4/3+4=(-4+12)/3=8/3
f(-1)=2/3(-1)+4=-2/3+4=(-2+12)/3=10/3
f(0)=2/3(0)+4=4
f(1)=2/3(1)+4=2/3+4=(2+12)/3=14/3
f(2)=2/3(2)+4=4/3+4=(4+12)/3=16/3
{8/3,10/3,4,14/3,16/3}
you're can graph those coordinates
actually you can take other coordinates...
CMIIW
,
Suppose we wish to determine whether or not two given polynomials with complex coefficients have a common root. Given two first-degree polynomials a0 + a1x and b0 + b1x, we seek a single value of x such that
Solving each of these equations for x we get x = -a0/a1 and x = -b0/b1 respectively, so in order for both equations to be satisfied simultaneously we must have a0/a1 = b0/b1, which can also be written as a0b1 - a1b0 = 0. Formally we can regard this system as two linear equations in the two quantities x0 and x1, and write them in matrix form as
Hence a non-trivial solution requires the vanishing of the determinant of the coefficient matrix, which again gives a0b1 - a1b0 = 0.
Now consider two polynomials of degree 2. In this case we seek a single value of x such that
Hope this helped, Hope I did not make it to complated
Please give me Brainliest
Answer: (x, y) = (-5, -10)
Step-by-step explanation: too lazy
K-7.2>2.1.
+7.2 +7.2
------------------
k> 9.3
are you suppose to solve for k?
Answer:
24
Step-by-step explanation:
definition of coefficient: a numerical or constant quantity placed before and multiplying the variable in an algebraic expression (e.g. 4 in 4x y)
