Answer:
x = 80°
Step-by-step explanation:
Angle on a straight line = 180°
(x + 40)° + (x - 20)° = 180°
x + 40° + x - 20° = 180°
2x + 20° = 180°
2x = 180° - 20°
2x = 160°
x = 160°/2
x = 80°
Check:
(x + 40)° + (x - 20)° = 180°
(80° + 40°) + (80° - 20°) = 180°
120° + 60° = 180°
180° = 180°
Answer:
The pairs of integer having two real solution for
are




Step-by-step explanation:
Given

Now we will solve the equation by putting all the 6 pairs so we get the following
for 
for 
for 
for 
for 
for 
The above all are Quadratic equations inn general form 
where we have a,b and c constant values
So for a real Solution we must have

for
we have
which is less than 0 ∴ not a real solution.
for
we have
which is greater than 0 ∴ a real solution.
for
we have
which is greater than 0 ∴ a real solution.
for
we have
which is greater than 0 ∴ a real solution.
for
we have
which is equal to 0 ∴ a real solution.
for
we have
which is less than 0 ∴ not a real solution.
What if I write the equation like this:
y = zero x - 3.
Now you know that the slope of the line is zero and its y-intercept is -3.
That's a horizontal line that crosses the y-axis at -3. And just like the equation says, the value of 'y' doesn't depend on 'x'. It's -3 everywhere.
Answer:
r = t - u - s
Step-by-step explanation:
To solve for r in terms of s, t, and u, means to have r on one side and have t, u, and s.
t = u - s + r
t - u = s + r
t - u - s = r
Pick any two points
(0, 3), (-1,9)
Find slope (y2-y1)/(x2-x1)
(9-3)/(-1-0) = 6/-1 = -6
Y = -6x + b
Plug in a point
3 = 0 + b, b= 3
Final equation: y = -6x + 3