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Mandarinka [93]
2 years ago
10

Which expression is a factor of both x2-3x+2 and

Mathematics
1 answer:
Margarita [4]2 years ago
6 0

Answer:

Step-by-step explanation:

So, factor each of the expressions and find out.  Why should anyone else factor the expressions for you?  Come on and don't be so lazy!

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Find the zeros : x(x^2+13x+40)
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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.
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ASAP!!! Pls help me out :(
mr_godi [17]

The answer you are looking for is B. 70.69

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Choose the graph that represents the equation y=3[x+1]+1.
MatroZZZ [7]

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

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  • It lowest point is -1,1
  • y never equals zero
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7 0
2 years ago
What is the domain of this graph?
professor190 [17]

Answer:

B

Step-by-step explanation:

The answer is b because the domain of anything is always x or on the x axis .

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3 years ago
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1. Let a; b; c; d; n belong to Z with n > 0. Suppose a congruent b (mod n) and c congruent d (mod n). Use the definition
lukranit [14]

Answer:

Proofs are in the explantion.

Step-by-step explanation:

We are given the following:

1) a \equi b (mod n) \rightarrow a-b=kn for integer k.

1) c \equi  d (mod n) \rightarrow c-d=mn for integer m.

a)

Proof:

We want to show a+c \equiv b+d (mod n).

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 a+c \equiv b+d (mod n).

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, a+c \equiv b+d (mod n).

//

b) Proof:

We want to show ac \equiv bd (mod n).

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 ac \equiv bd (mod n).

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, ac \equiv bd (mod n).

//

3 0
3 years ago
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