The zeros are x= 0,-5,-8. the zeros are the x values where the graph intersects the x-axis. to find the zeros, you have to replace y with zero then solve for x.
The answer you are looking for is B. 70.69
Answer:
Step-by-step explanation:
Discussion
You haven't given us any choices. Always do that it you have them. Look for the graph that looks like the one below. It has the following characteristics
- It opens upward
- It lowest point is -1,1
- y never equals zero
- The y intercept is 0,4
Answer:
B
Step-by-step explanation:
The answer is b because the domain of anything is always x or on the x axis .
Answer:
Proofs are in the explantion.
Step-by-step explanation:
We are given the following:
1)
for integer
.
1)
for integer
.
a)
Proof:
We want to show
.
So we have the two equations:
a-b=kn and c-d=mn and we want to show for some integer r that we have
(a+c)-(b+d)=rn. If we do that we would have shown that
.
kn+mn = (a-b)+(c-d)
(k+m)n = a-b+ c-d
(k+m)n = (a+c)+(-b-d)
(k+m)n = (a+c)-(b+d)
k+m is is just an integer
So we found integer r such that (a+c)-(b+d)=rn.
Therefore,
.
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b) Proof:
We want to show
.
So we have the two equations:
a-b=kn and c-d=mn and we want to show for some integer r that we have
(ac)-(bd)=tn. If we do that we would have shown that
.
If a-b=kn, then a=b+kn.
If c-d=mn, then c=d+mn.
ac-bd = (b+kn)(d+mn)-bd
= bd+bmn+dkn+kmn^2-bd
= bmn+dkn+kmn^2
= n(bm+dk+kmn)
So the integer t such that (ac)-(bd)=tn is bm+dk+kmn.
Therefore,
.
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