The intercepts and the graph on your worksheet are not correct. Please see below for details:
has solutions at x=-1 and x=3 (use the quadratic formula to solve). That means these are the x-intercepts, namely points:
(-1,0) and (3,0).
The y-intercept comes from setting x=0 and calculating the y value:
so the y-intercept is (0,-3).
Now to the graph: Based on the form of the function we can see this is a quadratic function and its graph will be a parabola. You can reformat the expression in the following form
and that will indicate that the apex of the parabola (open up) will be at the point (1,-4).
Knowing the apex, the x intercepts, and the y intercept, we can graph it now.
Graph is in the image attached.
To find angle Y, take 180 degrees (linear pair) minus 35 degrees (angle X).
180-35=145 degrees
Angle Y=145 degrees
Angle W is also 145 degrees since it’s a vertical angle to Angle Y.
To find angle N, take 90 degrees (complementary angle) minus 55 degrees (angle M).
90-55=35 degrees
Angle N=35 degrees
Hope this helped!
It is 18-6 out of all the other ones
Answer:
Step-by-step explanation:
A system of linear equations is one which may be written in the form
a11x1 + a12x2 + · · · + a1nxn = b1 (1)
a21x1 + a22x2 + · · · + a2nxn = b2 (2)
.
am1x1 + am2x2 + · · · + amnxn = bm (m)
Here, all of the coefficients aij and all of the right hand sides bi are assumed to be known constants. All of the
xi
’s are assumed to be unknowns, that we are to solve for. Note that every left hand side is a sum of terms of
the form constant × x
Solving Linear Systems of Equations
We now introduce, by way of several examples, the systematic procedure for solving systems of linear
equations.
Here is a system of three equations in three unknowns.
x1+ x2 + x3 = 4 (1)
x1+ 2x2 + 3x3 = 9 (2)
2x1+ 3x2 + x3 = 7 (3)
We can reduce the system down to two equations in two unknowns by using the first equation to solve for x1
in terms of x2 and x3
x1 = 4 − x2 − x3 (1’)
1
and substituting this solution into the remaining two equations
(2) (4 − x2 − x3) + 2x2+3x3 = 9 =⇒ x2+2x3 = 5
(3) 2(4 − x2 − x3) + 3x2+ x3 = 7 =⇒ x2− x3 = −1
