Answer:
In the graph we can find two points, lets select:
(2, 15) and (4, 30)
Those are the first two points.
Now, for two pairs (x1, y1) (x2, y2)
The slope of the linear equation y = s*x + b that passes trough those points is:
s = (y2 - y1)/(x2 - x1)
So the slope for our equation is
s = (30 - 15)/(4 - 2) = 15/2
then our linear equation is
y = (15/2)*x + b
now we can find b by imposing that when x = 2, y must be 15 (for the first point we selected)
15 = (15/2)*2 + b = 15 + b
b = 15 - 15 = 0
then our equation is:
y = (15/2)*x
Where we used a division and a multiplication.
I guess irreducible form? (translated from my language, maybe it's actually another word?)
Answer:
x=0
Step-by-step explanation:
Subtract 1 from both sides
2x+1-1<1-1
Simplify the arithmetic
2x<1-1
Simplify the arithmetic
2x<0
Isolate the X
X=0
Yes, it is possible as shown in the picture attached. There are two lines which are different functions. They intersect at one x-intercept. As the two lines extend infinitely at both ends, their domains and ranges are from ∞ to +∞. Therefore, the two functions drawn as two lines, have an identical x-intercept, domain and range.
X + y = 39
xy = -880
x + y = 39
x - x + y = -x + 39
y = -x + 39
xy = -880
x(-x + 39) = -880
x(-x) + x(39) = -880
-x² + 39x = -880
-x² + 39x + 880 = 0
-1(x²) - 1(-39x) - 1(-880) = 0
-1(x² - 39x - 880) = 0
-1 -1
x² - 39x - 880 = 0
x² - 55x + 16x - 880 = 0
x(x) - x(55) + 16(x) - 16(55) = 0
x(x - 55) + 16(x - 55) = 0
(x + 16)(x - 55) = 0
x + 16 = 0 or x - 55 = 0
- 16 - 16 + 55 + 55
x = -16 or x = 55
x + y = 39
-16 + y = 39
+ 16 + 16
y = 55
(x, y) = (-16, 55)
or
x + y = 39
55 + y = 39
- 55 - 55
y = -16
(x, y) = (55, -16)
The two numbers that add up to 39 and multiply to -880 are 55 and -16.