I'm assuming that x is part of the data set, and, with x, the mean equals 105. To find the value of x, you must add all the data values together to get 544+x (you still don't know what x equals). Then put 544+x over how many days values there are, including x (there are 6). You should have 544+x/6. Now, as this is how you would calculate the mean if you knew what x was equal to, you must set it equal to the mean, since you know what it is (105). You should now have 544+x/6 = 105. You have your equation set up--now you just have to solve it. I would multiply by 6 on both sides to get rid of the 6 on the left side. You would then have 544+x = 630. I would finally subtract 544 from both sides to get x = 86. Your final answer is x = 86.
G(x) = 2x² - 5x + 2 = 2x² - 4x - x + 2 = 2x · x - 2x · 2 - 1 · x - 1 · (-2)
= 2x(x - 2) -1(x - 2) = (x - 2)(2x - 1)
g(x) = 0 ⇔ (x - 2)(2x - 1) = 0 ⇔ x - 2 = 0 or 2x - 1 = 0
x = 2 or x = 0.5
Answer:
a) 13 m/s
b) (15 + h) m/s
c) 15 m/s
Step-by-step explanation:
if the location is
y=x²+3*x
then the average velocity from 3 to 7 is
Δy/Δx=[y(7)-y(3)]/(7-3)=[7²+3*7- (3²+3*3)]/4= 13 m/s
then the average velocity from x=6 to to x=6+h
Δy/Δx=[y(6+h)-y(6)]/(6+h-6)=[(6+h)²+3*(6+h)- (6²+3*6)]/h= (2*6*h+3*h+h²)/h=2*6+3= (15 + h) m/s
the instantaneous velocity can be found taking the limit of Δy/Δx when h→0. Then
when h→0 , limit Δy/Δx= (15 + h) m/s = 15 m/s
then v= 15 m/s
also can be found taking the derivative of y in x=6
v=dy/dx=2*x+3
for x=6
v=dy/dx=2*6+3 = 12+3=15 m/s
The ball will be at height 19 feet first at 0.7 sec and then at 1.42 sec.
<u>Explanation:</u>
We need to find the time at which the ball will be at height 19 feet.
Equation:
h = 3 + 34t - 16t²
19 = 3 + 34t - 16t²
16 = -16t² + 34t
-16t² + 34t - 16 = 0
On solving the equation, we get
t1 = 0.7 s and t2 = 1.42s
Therefore, the ball will be at height 19 feet first at 0.7 sec and then at 1.42 sec.
Answer:
look for the missing angle
