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makvit [3.9K]
3 years ago
14

Who can get this if you can ill give you an brain thingy and i do need help on this

Mathematics
2 answers:
tigry1 [53]3 years ago
4 0
The answer is abovisley 0 or 1
Dmitry [639]3 years ago
3 0
The answer is 12!!!!!!!!!!!
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If P(x)= 3x^4 - 5x^3 - 17x^2 + 13x + 6, and P(1)=0, and P(-2)=0, then the factorization of P(x) is?
liberstina [14]

Answer:

P(x) = (x - 1)(x + 2)(x - 3)(3x + 1)

Step-by-step explanation:

Since P(1) = 0 and P(- 2) = 0, then

(x - 1) and (x + 2) are factors of P(x)

(x - 1)(x + 2) = x² + x - 2 ← is also a factor of P(x)

dividing 3x^{4} - 5x³ - 17x² + 13x + 6 by x² + x - 2 gives

P(x) = (x - 1)(x + 2)(3x² - 8x - 3) = (x - 1)(x + 2)(x - 3)(3x + 1)


5 0
3 years ago
500=100(x-1)<br> Someone tell me step by step
Andre45 [30]
First, distribute the 100 to the (x-1) portion:
500 = 100x - 100

then, add 100 to both sides of your equation to get “x” alone on the right side:
500 + 100 = 100x - 100 + 100
600 = 100x

next, divide 100x by 100 and 600 by 100 to isolate a singular “x”:
600/100 = 100x/100
6 = x

therefore, x=6 is your solution
5 0
2 years ago
Help ASAP please
Neko [114]
C. Is the best answer
7 0
3 years ago
Please help me , i don't know this it's 7th grade math
scZoUnD [109]

Answer:

C. 1920

Step-by-step explanation:

25*8*10

1920

8 0
2 years ago
Read 2 more answers
1. Let a; b; c; d; n belong to Z with n &gt; 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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