Answer:
<em><u>x=-6, y=-2</u></em>, (As a point) (-6, -2).....The point form is not necessary unless you want to solve the system (of equations) by graphing.
Step-by-step explanation:
By substitution:
x-y=-4 By adding y on both sides,
x=y-4
Now you can substitute x for the expression (y-4)
Plug the (y-4) as x in the other equation.
So -2x+3y=6 becomes
-2(y-4)+3y=6
Now solve:
-2(y-4)+3y=6 distributes out to be
-2y+8+3y=6 Now combining like terms
y+8=6 Subtract 8 on both sides to isolate the variable
<u><em>y=-2</em></u>
Now plug the value -2 in where the y is in any equation (preferably the easier/less complicated one) and solve for x.
So x-y=-4 becomes
x-(-2)=-4
=x+2=-4
=<u><em>x=-6</em></u>
Answer:
y = - 7
Step-by-step explanation:
The equation of a horizontal line is
y = c
where c is the value of the y- coordinates the line passes through.
The line passes through (5, - 7) with y- coordinate - 7, thus
y = - 7 ← equation of horizontal line
Answer:
The standard form is 8 y ⁵ - 17 y⁴ + 6 y³ +2 y² - 11
The degree of given polynomial is '5'
the co-efficient of y⁴ is '-17'
Step-by-step explanation:
Given standard form 2 y²+ 6 y³-11-17 y⁴+8 y⁵
<em>The form ax² + b x + c is called the standard form of the quadratic expression of 'x'.This is second degree standard form of polynomial.</em>
<em>The form ax⁵ + b x⁴ + c x³ +d x² +ex +f is called the standard form of the quadratic expression of 'x'.This is fifth degree standard form of polynomial</em>
now Given polynomial is 2 y²+ 6 y³-11-17 y⁴+8 y⁵
The standard form is
8 y ⁵ - 17 y⁴ + 6 y³ +2 y² - 11
<u><em>Conclusion</em></u>:-
<em>The degree of given polynomial is '5'</em>
<em>The co-efficient of y⁴ is '-17'</em>
<em> </em>
AAS because there’s a vertical angle
