59 = j + 29
59-29=j+29-29
30 = j
(or you could rewrite it as j=30)
hope this helps!! :)
<h2>
Answer:</h2>
y =
x + 3
<h2>
Step-by-step explanation:</h2>
As shown in the graph, the line is a straight line. Therefore, the general equation of a straight line can be employed to derive the equation of the line.
The general equation of a straight line is given by:
y = mx + c <em>or </em>-------------(i)
y - y₁ = m(x - x₁) -----------------(ii)
Where;
y₁ is the value of a point on the y-axis
x₁ is the value of the same point on the x-axis
m is the slope of the line
c is the y-intercept of the line.
Equation (i) is the slope-intercept form of a line
Steps:
(i) Pick any two points (x₁, y₁) and (x₂, y₂) on the line.
In this case, let;
(x₁, y₁) = (0, 3)
(x₂, y₂) = (4, -2)
(ii) With the chosen points, calculate the slope <em>m</em> given by;
m = 
m = 
m = 
(iii) Substitute the first point (x₁, y₁) = (0, 3) and m =
into equation (ii) as follows;
y - 3 =
(x - 0)
(iv) Solve for y from (iii)
y - 3 =
x
y =
x + 3 [This is the slope intercept form of the line]
Where the slope is
and the intercept is 3
Answer:
None of those answers are correct
Step-by-step explanation:
There are no real solutions
Let's solve your equation step-by-step.
2+2x−3
2x+3
=
3x+4
x+2
Step 1: Cross-multiply.
2+2x−3
2x+3
=
3x+4
x+2
(2+2x−3)*(x+2)=(3x+4)*(2x+3)
2x2+3x−2=6x2+17x+12
Step 2: Subtract 6x^2+17x+12 from both sides.
2x2+3x−2−(6x2+17x+12)=6x2+17x+12−(6x2+17x+12)
−4x2−14x−14=0
For this equation: a=-4, b=-14, c=-14
−4x2+−14x+−14=0
Step 3: Use quadratic formula with a=-4, b=-14, c=-14.
x=
−b±√b2−4ac
2a
x=
−(−14)±√(−14)2−4(−4)(−14)
2(−4)
x=
14±√−28
−8
Answer:
No real solutions.
4 terms:
2x^2,5x,7y, and -6.