Jason has 5 boxes and 3 cases of soup cans Each box contains 6 cans and each case contains 36 cans How many cans of soup does Ja
1 answer:
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So even postive integers are by defention in form 2k where k is a natural number so
let the sum of even integers to n=S
S=2(1+2+3+4+5+6+7+8+......+k-1+k
divide bith sides of equation 1 by 2
0.5S=1+2+3+4+5+...........+k-1+k
S=2(k+(k-1)+..............................+2+1)
divide both sides of equation 2 by 2
0.5S=k+k-1+..............................+2+1)
by adding both we will get
___________________________
S=(k+1)(k)
so the sum will be equal to
S=
so let us test the equation
for the first 3 even number there sums will be
2+4+6=12
by our equation 3^2+3=12
gave us the same answer so our equation is correct
let the sum of even integers to n=S
S=2(1+2+3+4+5+6+7+8+......+k-1+k
divide bith sides of equation 1 by 2
0.5S=1+2+3+4+5+...........+k-1+k
S=2(k+(k-1)+..............................+2+1)
divide both sides of equation 2 by 2
0.5S=k+k-1+..............................+2+1)
by adding both we will get
___________________________
S=(k+1)(k)
so the sum will be equal to
S=
so let us test the equation
for the first 3 even number there sums will be
2+4+6=12
by our equation 3^2+3=12
gave us the same answer so our equation is correct
The function is 
.
To find the x-intercepts, we need to factorize the function. A very good idea is to first try the factors of 36:
f(1)=1-11+36-36, not 0
f(2)=8-44+72-36=-36+72-36=0. Here we have our first root (2).
Now we cad divide f(x) by (x-2) which will give us a quadratic expression, which we can factorize easily (if the discriminant is non negative).
We can also try some other factors of 36. Indeed we can check that
f(3)=27-99+108-36=135-135=0,
and
f(6)=216-396+216-36=0.
Thus, f(x)=(x-2)(x-3)(x-6). Note that if we expanded the right hand side expression, the constant term would be the product of the constants 2, 3, 6.
This is the reason why in the first place we looked at the factors of 36 for the possible zeros of f(x).
Thus, the x-intercepts are (2, 0), (3, 0), (6, 0), or 2, 3, 6.
The y-intercept is f(0), which is -36.
Note that f(0)<f(2) because f(0)=-36 and f(2)=0. This means that at the left side, the graph is coming from - infinity. Similarly,
we can check that f(10)=1000-1100+360-36=224> f(6). That is, to the right of our rightmost root, the graph is getting larger.
Thus, the end behaviors are: the graph goes to + infinity as x goes to + infinity,
and it goes to minus infinity as x goes to - infinity.
To find the x-intercepts, we need to factorize the function. A very good idea is to first try the factors of 36:
f(1)=1-11+36-36, not 0
f(2)=8-44+72-36=-36+72-36=0. Here we have our first root (2).
Now we cad divide f(x) by (x-2) which will give us a quadratic expression, which we can factorize easily (if the discriminant is non negative).
We can also try some other factors of 36. Indeed we can check that
f(3)=27-99+108-36=135-135=0,
and
f(6)=216-396+216-36=0.
Thus, f(x)=(x-2)(x-3)(x-6). Note that if we expanded the right hand side expression, the constant term would be the product of the constants 2, 3, 6.
This is the reason why in the first place we looked at the factors of 36 for the possible zeros of f(x).
Thus, the x-intercepts are (2, 0), (3, 0), (6, 0), or 2, 3, 6.
The y-intercept is f(0), which is -36.
Note that f(0)<f(2) because f(0)=-36 and f(2)=0. This means that at the left side, the graph is coming from - infinity. Similarly,
we can check that f(10)=1000-1100+360-36=224> f(6). That is, to the right of our rightmost root, the graph is getting larger.
Thus, the end behaviors are: the graph goes to + infinity as x goes to + infinity,
and it goes to minus infinity as x goes to - infinity.