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3 years ago

Translate mathemtically : an even number

1 answer:

7 0

-- To create an even number:

   2n is always even, no matter what 'n' is.

-- To test a number and see whether it's even:

   'n' is even if (n/2) = (the greatest integer in n/2)

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I’m kind of lazy to answer this question can somebody answer this for me

kirill [66]

Answer:

everyone got 6 pieces

Step-by-step explanation:

do you actually need a explanation?

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3 years ago

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Determine if Ginny factored correctly. If not, explain where she made an error. Ginny made a mistake in step 1 when she identifi

il63 [147K]

Quadratic equations are second-order equations. Ginney made a mistake when she identified b = 31. It should be b = –31.

<h3>What is a Quadratic Equation?</h3>

A quadratic equation is an equation that can be written in the form of

ax²+bx+c.

Where a is the leading coefficient, and

c is the constant.

In order to find the mistake that Ginny made, we will first find the factors ourselves, therefore, we will factorize the trinomial,

$6x^2 - 31x -30 = 0$

As we can see that the value of constants are,

a = 6

b = -31

c = 30

$6x^2 - 31x -30 = 0\\\\6x^2 - 36x+5x -30 = 0\\\\6x(x-6)+5(x-6)=0\\\\(6x+5)(x-6)=0$

Therefore, the factors of the trinomial $6x^2 - 31x -30 = 0$ are (6x+5) and (x-6).

If we compare our process with Ginny's process, we will find that she has taken the value of the constant b as 31 which is wrong.

Hence, Ginney made a mistake when she identified b = 31. It should be b = –31.

Learn more about Quadratic Equation:

brainly.com/question/17177510

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3 years ago

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Prove that u(n) is a group under the operation of multiplication modulo n.

katrin2010 [14]

Answer:

The answer is the proof so it is long.

The question doesn't define u(n), but it's not hard to guess.

Group G with operation ∘

For all a and b and c in G:

1) identity: e ∈ G, e∘a = a∘e = a,

2) inverse: a' ∈ G, a∘a' = a'∘a = e,

3) closed: a∘b ∈ G,

4) associative: (a∘b)∘c = a∘(b∘c),

5) (optional) commutative: a∘b = b∘a.

Define group u(n) for n prime is the set of integers 0 < i < n with operation multiplication modulo n.

If n isn't prime, we exclude from the group all integers which share factors with n.

Identity: e = 1. Clearly 1∘a = a∘1 = a. (a is already < n).

Closed: u(n) is closed for n prime. We must show that for all a, b ∈ u(n), the integer product ab is not divisible by n, so that ab ≢ 0 (mod n). Since n is prime, ab ≠ n. Since a < n, b < n, no factors of ab can equal prime n. (If n isn't prime, we already excluded from u(n) all integers sharing factors with n).

Inverse: for all a ∈ u(n), there is a' ∈ u(n) with a∘a' = 1. To find a', we apply Euclid's algorithm and write 1 as a linear combination of n and a. The coefficient of a is a' < n.

Associative and Commutative:

(a∘b)∘c = a∘(b∘c) because (ab)c = a(bc)

a∘b = b∘a because ab = ba.

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3 years ago

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Choose the function that correctly identifies the transformation of f(x) = x2 shifted two units to the left and one unit down.

AnnZ [28]

Answer:

D) g(x) = (x + 2)^2 - 1

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3 years ago