13/4 =12/4 +1/4=3+1/4 so it has to be between 3 and 4 on the number line.Hope this helps and thank you for letting me help you ✅
Short AnswerThere are two numbers
x1 = -0.25 + 0.9682i <<<<
answer 1x2 = - 0.25 - 0.9582i <<<<
answer 2 I take it there are two such numbers.
Let one number = x
Let one number = y
x + y = -0.5
y = - 0.5 - x (1)
xy = 1 (2)
Put equation 1 into equation 2
xy = 1
x(-0.5 - x) = 1
-0.5x - x^2 = 1 Subtract 1 from both sides.
-0.5x - x^2 - 1 = 0 Order these by powers
-x^2 - 0.5x -1 = 0 Multiply though by - 1
x^2 + 0.5x + 1 = 0 Use the quadratic formula to solve this.

a = 1
b = 0.5
c = 1

x = [-0.5 +/- sqrt(0.25 - 4)] / 2
x = [-0.5 +/- sqrt(-3.75)] / 2
x = [-0.25 +/- 0.9682i
x1 = -0.25 + 0.9682 i
x2 = -0.25 - 0.9682 i
These two are conjugates. They will add as x1 + x2 = -0.25 - 0.25 = - 0.50.
The complex parts cancel out. Getting them to multiply to 1 will be a little more difficult. I'll do that under the check.
Check(-0.25 - 0.9682i)(-0.25 + 0.9682i)
Use FOIL
F:-0.25 * -0.25 = 0.0625
O: -0.25*0.9682i
I: +0.25*0.9682i
L: -0.9682i*0.9682i = - 0.9375 i^2 = 0.9375
NoticeThe two middle terms (labled "O" and "I" ) cancel out. They are of opposite signs.
The final result is 0.9375 and 0.0625 add up to 1
Answer:
The graph representing f(x)=6cos(4πx) is shown in the attached file.
Step-by-step explanation:
The graph representing f(x)=6cos(4πx) is shown in the attached file.
a. Use the mean value theorem. 16 falls between 12 and 20, so

(Don't forget your units - 5 m/min^2)
b.
gives the Johanna's velocity at time
. The magnitude of her velocity, or speed, is
. Integrating this would tell us the total distance she has traveled whilst jogging.
The Riemann sum approximates the integral as

If you're not sure how this is derived: we're given 5 sample points, so we can cut the interval [0, 40] into 4 subintervals. The lengths of each subinterval are 12, 8, 4, and 16 (the distances between each sample point), and the height of the rectangle approximating the area under the plot of
is determined by the value of
at each sample point, 200, 240, |-220| = 220, and 150.
c. Bob's velocity is given by
, so his acceleration is given by
. We have

and at
his acceleration is
m/min^2.
d. Bob's average velocity over [0, 10] is given by the difference quotient,
m/min