10 pens=£6.50
1 pen=£0.65
4 pens=£0.65×4
4 pens=£2.6
Graphing is one way to do the problem.But sometimes, graphing it is hard to do.So here’s an algebraic method.
If M(m1, m2) is the midpoint of two points A(x1, y1) and B(x2, y2),then m1 = (x1 + x2)/2 and m2 = (y1 + y2)/2.In other words, the x-coordinate of the midpointis the average of the x-coordinates of the two points,and the y-coordinate of the midpointis the average of the y-coordinates of the two points.
Let B have coordinates (x2, y2) in our problem.Then we have that 6 = (2 + x2)/2 and 8 = (3 + y2)/2.
Solving for the coordinates gives x2 = 10, y2 = 13
Answer:
Henry's balloon was farther from the town at the beginning and Henry's balloon traveled more quickly.
Step-by-step explanation:
The distance of Tasha's balloon from the town is represented by the function y = 8x+ 20 ............. (1)
Where y is the distance in miles from the town and x represents the time of fly in hours.
So, at the start of the journey i.e. at x = 0, y = 20 miles {From equation (1)} from the town.
Again, Tasha's balloon is traveling at a rate of 8 miles per hour.
Now, Henry's balloon begins 30 miles from the town and is 48 miles from the town after 2 hours.
So, Henry's balloon traveling with the speed of 
miles per hour.
Therefore, Henry's balloon was farther from the town at the beginning i.e. 30 miles from the town. And Henry's balloon traveled more quickly i.e at the rate of 9 miles per hour. (Answer)
One A
y = e^x
dy/dx = e^x The f(x) = the differentiated function. Any value that e^x can have, the derivative has the same value. x is contained in all the reals.
One B
y = x*e^x
y' = e^x + xe^x Using the multiplication rule.
You want the slope and the value of the of y to be the same. The slope is y' of the tangent line
xe^x = e^x + xe^x
e^x = 0
This happens only when x is very "small" like x = - 4444444
y = e^x * ln(x) Using the multiplication rule again, we need the slope of the line with is y'
y1 = e^x
y1' = e^x
y2 = ln(x)
y2' = 1/x
y' = e^x*ln(x) + e^x/x So at x = 1 the slope of the line =
y' = e^1*ln(1) + e^1/1
y' = e*0+e = e
y = mx + b
y = ex + b
to find b we use y= e^x ln(x)
e^x ln(x) = e*x + b
e^1 ln(1) = e*1 + b
ln(1) = 0
0 = e + b
b = - e
line equation and answer.
y = e*x - e
Answer:
m³n
Step-by-step explanation:
