Answer:
60
Step-by-step explanation:
Given: 60% out of 100
Expressed mathematically:
= (60/100) x 100
= 60
substitution and elimination means that you need to use one equation and substitute it in place of some variable in the other equation.
Consider the above two equations.
Take equation 1, which is 4x-2y=22 => x=(22+2y)/4
substitute the x value gained above in equation 2.
so, 2((22+2y)/4)+4y=6
22+2y+8y=12 => 10y = -10 => y= -1.
Substitute y= -1 in x value obtained in the beginning.
So, x= (22 - 2)/4 => 5.
There fore, x= 5 and y= -1
Hope it helps.
Let 4 + 3x = u
then 3dx = du
or, <span>1/3∫<span>u√</span>du
</span>
= <span>1/3<span>u^<span>3/2</span></span>/(3/2)
</span>
= <span>2/9∗(4+3x<span>)^<span>3/2
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Answer:
The rate at which the distance between them is changing at 2:00 p.m. is approximately 1.92 km/h
Step-by-step explanation:
At noon the location of Lan = 300 km north of Makenna
Lan's direction = South
Lan's speed = 60 km/h
Makenna's direction and speed = West at 75 km/h
The distance Lan has traveled at 2:00 PM = 2 h × 60 km/h = 120 km
The distance north between Lan and Makenna at 2:00 p.m = 300 km - 120 km = 180 km
The distance West Makenna has traveled at 2:00 p.m. = 2 h × 75 km/h = 150 km
Let 's' represent the distance between them, let 'y' represent the Lan's position north of Makenna at 2:00 p.m., and let 'x' represent Makenna's position west from Lan at 2:00 p.m.
By Pythagoras' theorem, we have;
s² = x² + y²
The distance between them at 2:00 p.m. s = √(180² + 150²) = 30·√61
ds²/dt = dx²/dt + dy²/dt
2·s·ds/dt = 2·x·dx/dt + 2·y·dy/dt
2×30·√61 × ds/dt = 2×150×75 + 2×180×(-60) = 900
ds/dt = 900/(2×30·√61) ≈ 1.92
The rate at which the distance between them is changing at 2:00 p.m. ds/dt ≈ 1.92 km/h
