Answer:
7.18 light-years
Step-by-step explanation:
6.8 x 10^13 = 68,000,000,000,000
9.46 x10 ^12 = 9,460,000,000,000
6.8 x 10^13 * 1 light year / 9.46 x 10^12 = 7.188 light years.
Have a Merry X-MAS Plz MArk Branilest
Answer:
t≈8.0927
Step-by-step explanation:
h(t) = -16t^2 + 128t +12
We want to find when h(t) is zero ( or when it hits the ground)
0 = -16t^2 + 128t +12
Completing the square
Subtract 12 from each side
-12 = -16t^2 + 128t
Divide each side by -16
-12/-16 = -16/-16t^2 + 128/-16t
3/4 = t^2 -8t
Take the coefficient of t and divide it by 8
-8/2 = -4
Then square it
(-4) ^2 = 16
Add 16 to each side
16+3/4 = t^2 -8t+16
64/4 + 3/4= (t-4)^2
67/4 = (t-4)^2
Take the square root of each side
±sqrt(67/4) =sqrt( (t-4)^2)
±1/2sqrt(67) = (t-4)
Add 4 to each side
4 ±1/2sqrt(67) = t
The approximate values for t are
t≈-0.092676
t≈8.0927
The first is before the rocket is launched so the only valid answer is the second one
Answer:
the answer is 7.2 10³
Step-by-step explanation:
7.2 10³
The answer is: "2.5 years" .
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Note: I = P * r * t ; { " Interest = Principal * rate * time "} ;
→ Solve for "t" {"time", in years} ;
Divide each side of the equation by "{P * r}" ;
to isolate "t" on one side of the equation ;
→ I / (P * r) = {P * r * t) / (P * r} ;
to get: " I / (P * r) = t " ;
↔ t = I / (P * r) ;
Given: I = $450 ;
<span>P = $2400 ;
r = 7.5% = 7.5/100 = 0.075 ;
Plug in these values into the formula to solve for the time, "t" :
</span>→ t = I / (P * r ) ;
= $450 / (<span>$2400 * 0.075) ;
= </span>$450 / ($2400 * 0.075) ;
= $450 / $180 ;
= $45 / $18 ;
= ($45 ÷ 9) / ($18 ÷ 9)
= $5 / $2 ;
= 2.5 ;
→ t = 2.5 years.
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The answer is: "2.5 years" .
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