Answer:
The answer is True
Step-by-step explanation:
A <em>mathematical induction</em> consists in only 2 steps:
<u>First step</u>: Show the proposition is true for the first one valid integer number.
<u>Second step</u>: Show that if any one is true then the next one is true
Finally, if first step and second step are true, then the complete proposition is true.
So, given 
First step: using and replacing n=2 (the first valid integer number >1)


As the result is an integer number, so the first step is true.
Second step: using any next number,
, let it replace

As the First step is true, we know that

,
So let it replace in the previous expression
![6*k+6*(n^2+2*n+1)\\6*[k+(n^2+2*n+1)]](https://tex.z-dn.net/?f=6%2Ak%2B6%2A%28n%5E2%2B2%2An%2B1%29%5C%5C6%2A%5Bk%2B%28n%5E2%2B2%2An%2B1%29%5D)
Finally
![\frac{6*[k+(n^2+2*n+1)]}{6} =k+(n^2+2*n+1)](https://tex.z-dn.net/?f=%5Cfrac%7B6%2A%5Bk%2B%28n%5E2%2B2%2An%2B1%29%5D%7D%7B6%7D%20%3Dk%2B%28n%5E2%2B2%2An%2B1%29)
where the last expression is an integer number
So the second step is true, and the complete proposition is True
Answer:
PΔJKL=66
Step-by-step explanation:
so we are given the line segments JK, KL, and LJ which are tangent to k(O), and also that JA=9, AL=10, and CK=14
JL=JA+AL (parts whole postulate)
JL=9+10=19 (substitution, algebra)
JA=JB=9 (tangent segments from the same point are congruent)
CK=KB=14 (tangent segments from the same point are congruent)
JK=JB+KB (parts whole postulate)
JK=9+14=23 (substitution, algebra)
LA=LC=10 (tangent segments from the same point are congruent)
LK=LC+CK (parts whole postulate)
LK=10+14=24 (substitution, algebra)
Perimeter of ΔJKL=LK+KL+LJ (perimeter formula for triangles)
Perimeter of ΔJKL=23+24+19=66 (substitution, algebra)
Answer:
x = 11/10
Step-by-step explanation:
we have from statement:
-1=1 - 4(-5+4x) – 4x
we apply distributive property:
-1 = 1 -4*(-5) -4*(4x) -4x
so we have:
-1= 1+20 -16x -4x
-1 = 21 -20x
-22=-20x
so we have:
x=22/20
x=11/10
we can substitute the x value found:
-1 = 1 - 4(-5+4*(11/10)) – 4*(11/10)
-1 = 1+20 -17.6 -4.4
-1 =21 -22
so we have:
-1 = -1
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Formula
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Volume = πr²h
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Find Volume of one tank
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Volume of 1 tank = πr²h
Volume of 1 tank = 3.14 x 6² x 12
Volume of 1 tank = 678.24 ft³
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Find Volume of 5 tanks
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Volume of 1 tank = 678.24
Volume of 5 tanks = 678.24 x 5
Volume of 5 tanks = 3391.2 ft³
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Answer: 3391.2 ft³
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