<span>A foot per second is a unit of speed. Something traveling one foot per second is traveling 0.3048 meters per second, or about 0.682 miles per hour. hope this helped please mark me as the brainliest answer
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2x+5y=619
x=10y-3
________
2(10y-3) +5y=619
20y-6+5y=619
25y=619+6
25y=625
y=625/25
y=25
x=10*25-3
x=250-3
x=247
Answer:
26
Step-by-step explanation :
73 - 47 = 26 brainliest?
Answer:

Step-by-step explanation:
y′′ + 4y′ − 21y = 0
The auxiliary equation is given by
m² + 4m - 21 = 0
We solve this using the quadratic formula. So

So, the solution of the equation is

where m₁ = 3 and m₂ = -7.
So,

Also,

Since y(1) = 1 and y'(1) = 0, we substitute them into the equations above. So,


Substituting A into (1) above, we have

Substituting B into A, we have

Substituting A and B into y, we have

So the solution to the differential equation is

Answer:
z-4
Step-by-step explanation:
z(5z²-80) ÷ 5z(z+4) = z-4
Step-by-step explanation:
Given: z(5z²-80) ÷ 5z(z+4)
Calculation:
First we factor numerator z(5z²-80)
Take out 5 from 5z²-80 and we get 5z(z²-16)
Now we factor 16 = 4²
5z(z²-4²)
Using formula, (a²- b²)=(a+b)(a-b)
5z(z²- 4²)⇒5z(z+4)(z-4)
Simplified fraction
5z(z+4)(z-4) ÷ 5z(z+4)
Cancel like factor from numerator and denominator
⇒ z-4
Thus, z(5z²-80) ÷ 5z(z+4) = z-4