Answer:
(–∞, -4)
Step-by-step explanation:
Since both equations are equal to y, we can just combine them like this:
1/3x-3=-x+5
1/3x=-x+8
4/3x=8 (since x is 1x which is 3/3x)
x=6
Plug x back in:
y=-6+5
y=-1
So x=6 and y=-1
Hope this helped!
Answer:I believe it is (C) I’m not 100% sure but I think so
Step-by-step explanation:
Answer:
Step-by-step explanation:
x is greater or equal to 13
<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>
![\mathrm{H}=\frac{\left(\mathrm{v} 0^{2} \times \sin 2 \emptyset\right)}{2 \mathrm{g}}](https://tex.z-dn.net/?f=%5Cmathrm%7BH%7D%3D%5Cfrac%7B%5Cleft%28%5Cmathrm%7Bv%7D%200%5E%7B2%7D%20%5Ctimes%20%5Csin%202%20%5Cemptyset%5Cright%29%7D%7B2%20%5Cmathrm%7Bg%7D%7D)
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
![\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 2}{\mathrm{g} 1}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7BH%7D%201%7D%7B%5Cmathrm%7BH%7D%202%7D%3D%5Cfrac%7B%5Cmathrm%7Bg%7D%202%7D%7B%5Cmathrm%7Bg%7D%201%7D)
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>
![\mathrm{h}=(\mathrm{v} 0 * \mathrm{t})+\frac{\mathrm{a} *\left(t^{2}\right)}{2}](https://tex.z-dn.net/?f=%5Cmathrm%7Bh%7D%3D%28%5Cmathrm%7Bv%7D%200%20%2A%20%5Cmathrm%7Bt%7D%29%2B%5Cfrac%7B%5Cmathrm%7Ba%7D%20%2A%5Cleft%28t%5E%7B2%7D%5Cright%29%7D%7B2%7D)
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
![\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 1}{\mathrm{g} 2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7BH%7D%201%7D%7B%5Cmathrm%7BH%7D%202%7D%3D%5Cfrac%7B%5Cmathrm%7Bg%7D%201%7D%7B%5Cmathrm%7Bg%7D%202%7D)
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