M=5
Positive 5 because a negative divide by a negative is a positive
All you have to do is divide 45 and 9
Answer:
Constant
Step-by-step explanation:
For a straight line for example,
Take any two points from the line and find the slope,
it will be the same value for any 2 points chosen
So throughout the graph, the slope/rate of change is not changing
That is
Throughout the graph rate of change/slope is constant
There are two ways to do this
Method 1:
Find (f-g)(x) first
(f-g)(x) = f(x) - g(x)
(f-g)(x) = (5x^2+3) - (-2x+4)
(f-g)(x) = 5x^2+3+2x-4
(f-g)(x) = 5x^2+2x-1
Then plug in x = -3
(f-g)(-3) = 5(-3)^2+2(-3)-1
(f-g)(-3) = 5(9)+2(-3)-1
(f-g)(-3) = 45-6-1
(f-g)(-3) = 39-1
(f-g)(-3) = 38
-----------------------------------------
Method 2:
Find f(-3)
f(x) = 5x^2+3
f(-3) = 5(-3)^2+3
f(-3) = 5(9)+3
f(-3) = 45+3
f(-3) = 48
Find g(-3)
g(x) = -2x+4
g(-3) = -2(-3)+4
g(-3) = 6+4
g(-3) = 10
Subtract the two results
(f-g)(-3) = f(-3) - g(-3)
(f-g)(-3) = 48 - 10
(f-g)(-3) = 38
-----------------------------------------
Whichever method you pick, the answer is: 38
Answer:
The point C is 12.68 km away from the point A on a bearing of S23.23°W.
Step-by-step explanation:
Given that AB is 50 km and BC is 40 km as shown in the figure.
From the figure, the length of x-component of AC = |AB sin 80° - BC cos 20°|
=|50 sin 80° - 40 cos 20°|=11.65 km
The length of y-component of AC = |AB cos 80° - BC sin 20°|
=|50 cos 80° - 40 sin 20°|= 5 km
tan
= 5/11.65
=23.23°
AC= 
km
Hence, the point C is 12.68 km away from the point A on a bearing of S23.23°W.
Answer:
C
Step-by-step explanation:
I think the correct answer would be C since you are using lines and angles but I am not 100 percent sure. Also because A and B wouldn't make since because it is adding.
