Answer:
Step-by-step explanation:
diameter=2*radius
=2*9
=18 in
given central angle=72 degree
arc length=central angle /360° *πd
=72/360 *3.14*18
=1/5 *56.52
=56.52/5
=11.304
Answer:
d67d6d 34r2
Step-by-step explanation:
d53eududc6ttd5cf
Answer:
f(x) = -3x - 21
Step-by-step explanation:
Graphed this equation and it passes through (-6, -3), also known as Point A.
For this question, it would be most effective to use an algebraic expression to more easily show what the question is asking. If we use the variable "k" to show the distance in km that he cycled on Sunday, we know that the amount he cycled on Saturday equals k + 12, and the amount that he cycled on the weekend should be the amount of Saturday plus the amount of Sunday. If we write this as an equation we say:
k + k + 12 = 38
=> 2k + 12 = 38
Now we can just rearrange and solve for k:
=> 2k = 26
=> k = 26/2 = 13
Therefore Patrick cycled 13km on Sunday
To solve the answer, we just add 12km to the value for Sunday like so:
12 + 13 = the amount he cycled on Sunday
Hope this helped, remember to please try and understand the maths as well as the answer :))
<u>Answer:</u>
a) 3.675 m
b) 3.67m
<u>Explanation:</u>
We are given acceleration due to gravity on earth =
And on planet given =
A) <u>Since the maximum</u><u> jump height</u><u> is given by the formula </u>

Where H = max jump height,
v0 = velocity of jump,
Ø = angle of jump and
g = acceleration due to gravity
Considering velocity and angle in both cases

Where H1 = jump height on given planet,
H2 = jump height on earth = 0.75m (given)
g1 = 2.0
and
g2 = 9.8
Substituting these values we get H1 = 3.675m which is the required answer
B)<u> Formula to </u><u>find height</u><u> of ball thrown is given by </u>

which is due to projectile motion of ball
Now h = max height,
v0 = initial velocity = 0,
t = time of motion,
a = acceleration = g = acceleration due to gravity
Considering t = same on both places we can write

where h1 and h2 are max heights ball reaches on planet and earth respectively and g1 and g2 are respective accelerations
substituting h2 = 18m, g1 = 2.0
and g2 = 9.8
We get h1 = 3.67m which is the required height