Answer:
1 seconds after being thrown, the stone reaches its max height
Step-by-step explanation:
The parabolic (quadratic) equation is:
Lets expand this in the form 
, so we have:
We can say the values of a,b, and c, now to be:
a = -5
b = 10
c = 40
The number of seconds at which the max would occur is given by the point, x, at:
We know a and b, let's find the seconds, x,
Hence,
1 seconds after being thrown, the stone reach its max height
I am a number greater than 40,000 and less than 60,000:
40,000 < n < 60,000
This means that:
n = 10,000n₁ + 1,000n₂ + 100n₃ + 11n₄
And also:
4 ≤ n₁ < 6
0 ≤ n₂ ≤ 9
0 ≤ n₃ ≤ 9
0 ≤ n₄ ≤ 9
My ten thousands digit is 1 less than 3 times the sum of my ones digit and tens digit:
n₁ = 3*2n₄ - 1
n₁ = 6n₄ - 1
This means that:
n = 10,000*(6n₄-1) + 1,000n₂ + 100n₃ + 11n₄
n = 60,000n₄ - 10,000 + 1,000n₂ + 100n₃ + 11n₄
n = 60,011n₄ - 10,000 + 1,000n₂ + 100n₃
<span>My thousands digit is half my hundreds digit, and the sum of those two digits is 9:
n</span>₂ = 1/2 * n₃
<span>
n</span>₂ + n₃ = 9
<span>
Therefore:
n</span>₂ = 9 - n₃
<span>
Therefore:
9 - n</span>₃ = 1/2 * n₃
<span>
9 = 1/2 * n</span>₃ + n₃
<span>
9 = 1.5 * n</span>₃
<span>
Therefore:
n</span>₃ = 6
<span>
If n</span>₃=6, n₂=3.
<span>
This means that:
</span>n = 60,011n₄ - 10,000 + 1,000*3 + 100*6
n = 60,011n₄ - 10,000 + 3,000 + 600
n = 60,011n₄ - 6,400
Therefore:
0<n₄<2, so n₄=1.
If n₄=1:
n = 60,011 - 6,400
n = 53,611
Answer:
53,611
Answer:
Step-by-step explanation:
______________
Answer:
The length of the box is 10 feet.
Step-by-step explanation:
You have to work backwards to find volume, so just divide the volume (100) by the height (2) then divide that by the width (5) and you get your answer, 10
Answer:
see below
Step-by-step explanation:
(ab)^n=a^n * b^n
We need to show that it is true for n=1
assuming that it is true for n = k;
(ab)^n=a^n * b^n
( ab) ^1 = a^1 * b^1
ab = a * b
ab = ab
Then we need to show that it is true for n = ( k+1)
or (ab)^(k+1)=a^( k+1) * b^( k+1)
Starting with
(ab)^k=a^k * b^k given
Multiply each side by ab
ab * (ab)^k= ab *a^k * b^k
( ab) ^ ( k+1) = a^ ( k+1) b^ (k+1)
Therefore, the rule is true for every natural number n
