Answer:
no solution
Step-by-step explanation:
Suppose the system equations are:
y = (-1/4)x + 2 (1)
3y = (-3/4)x – 6 (2)
If we multiply equation (1) by 3:
3y = (-3/4)x + 6 (1)
3y = (-3/4)x – 6 (2)
then subtract it from equation (2):
0y = 0x – 12
0 = -12
This is not possible, therefore, the system equation has no solution
Answer:
Step-by-step explanation:
Only one
The radius of the circle is given to be r = 14.5
Let the center be O. So OB is the radius = 14.5
We can draw a triangle as shown in the image below. We have a right angled triangle. We know the hypotenuse and the base, we need to find the perpendicular side. This can be done using the Pythagorean theorem.
So, we can write:
![(14.5)^{2}=10^{2}+OA^{2} \\ \\ OA^{2}=110.25 \\ \\ OA=10.5](https://tex.z-dn.net/?f=%2814.5%29%5E%7B2%7D%3D10%5E%7B2%7D%2BOA%5E%7B2%7D%20%5C%5C%20%20%5C%5C%20%0AOA%5E%7B2%7D%3D110.25%20%5C%5C%20%20%5C%5C%20%0AOA%3D10.5%20%20%20%20)
Thus, the measure of OA = 10.5
OB = OA + AB
14.5 = 10.5 + AB
⇒
AB = 4
Answer:
Step-by-step explanation:
actually, it IS 7
U(x) = f(x).(gx)
v(x) = f(x) / g(x)
Use chain rule to find u(x) and v(x).
u '(x) = f '(x) g(x) + f(x) g'(x)
v ' (x) = [f '(x) g(x) - f(x) g(x)] / [g(x)]^2
The functions given are piecewise.
You need to use the pieces that include the point x = 1.
You can calculate f '(x) and g '(x) at x =1, as the slopes of the lines that define each function.
And the slopes can be calculated graphycally as run / rise of each graph, around the given point.
f '(x) = slope of f (x); at x = 1, f '(1) = run / rise = 1/1 = 1
g '(x) = slope of g(x); at x = 1, g '(1) = run / rise = 1.5/ 1 = 1.5
You also need f (1) = 1 and g(1) = 2
Then:
u '(1) = f '(1) g(1) + f(1) g'(1) = 1*2 + 1*1.5 = 2 + 1.5 = 3.5
v ' (x) = [f '(1) g(1) - f(1) g(1)] / [g(1)]^2 = [1*2 - 1*1.5] / (2)^2 = [2-1.5]/4 =
= 0.5/4 = 0.125
Answers:
u '(1) = 3.5
v '(1) = 0.125