Answer:
Step-by-step explanation:
d = 7%2F2
-2, -2+7%2F2, -2+2%287%2F2%29, -2+3%287%2F2%29, 12
-2, -4%2F2+7%2F2, -4%2F2+14%2F2, --4%2F2+21%2F2, 12
-2, 3%2F2, 10%2F2, 17%2F2, 12
-2, 3%2F2, 5, 17%2F2, 12
<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
Answer:
M∠DEC equals 123º.
Step-by-step explanation:
The sum of a triangle's three angles always equal 180º. The exterior angle, x, equals the two non-adjacent interior angles.
180 - {(x - 45)+(x - 12)} = m∠DEC
m∠DEC + x = 180
<u>(x - 45) + (x - 12) = x</u>
Solving for x:
(x - 45) + (x - 12) = x
x - 45 + x - 12 = x Remove parenthesis
2x - 57 = x Combine like terms
2x = x + 57 Add 57 to both sides
<u>x = 57</u><u> Subtract x from both sides</u>
Finding m∠D:
x - 45 = ?
<u>57 - 45 = </u><u>12º </u>
Finding m∠C:
x - 12 = ?
<u>57 - 12 = </u><u>45º </u>
<em>** </em><em>(Checking x: 12 + 45 = 57) </em><em>**</em>
<em>Finding </em>m∠DEC:
AC is a straight line, and because straight lines are equivalent to 180º, we subtract 57 from 180:
180 - 57 = 123º
Hope this helps,
❤<em>A.W.E.</em><u><em>S.W.A.N.</em></u>❤
Answer:
tan a + cot b
Step-by-step explanation:
It's already simplified.
There are alternate forms like
![sec(a)csc(b)cos(a-b)\\\\sec(a)csc(b)[sin(a)sin(b)+cos(a)cos(b)]\\\\\frac{sin (a)}{cos (a)} +\frac{cos(b)}{sin(b)}](https://tex.z-dn.net/?f=sec%28a%29csc%28b%29cos%28a-b%29%5C%5C%5C%5Csec%28a%29csc%28b%29%5Bsin%28a%29sin%28b%29%2Bcos%28a%29cos%28b%29%5D%5C%5C%5C%5C%5Cfrac%7Bsin%20%28a%29%7D%7Bcos%20%28a%29%7D%20%2B%5Cfrac%7Bcos%28b%29%7D%7Bsin%28b%29%7D)