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LuckyWell [14K]
3 years ago
8

What’s the answer to (16-x)2?

Mathematics
1 answer:
Oxana [17]3 years ago
6 0

Answer:

-x + 18

Step-by-step explanation:

just add 16 and 2. this is the furthest it can be simplified

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What is (2+y)³ written out? 8 + 12y + 6y² + y³ 6 + 8y + 4y² + y³
AveGali [126]
\boxed{(2+y)^3-cube\ the\ sum\ of\ the\ numbers\ 2\ and\ y}

Use:\ (a+b)^3=(a+b)(a+b)(a+b)=a^3+3a^2b+3ab^2+b^3\\----------------------------\\(2+y)^3=2^3+3\cdot2^2\cdot y+3\cdot2\cdot y^2+y^3=8+3\cdot4y+6y^2+y^3\\\\=8+12y+6y^2+y^3\\========================================

8+12y+6y^2+6y^3+8y+4y^2+y^3\\=8+(12y+8y)+(6y^2+4y^2)+(6y^3+y^3)\\=8+20y+10y^2+7y^3



7 0
3 years ago
Read 2 more answers
I need help in sovling this probelm. simplify (2xyz)^0
Zielflug [23.3K]
Anything raised to the 0 power = 1
5 0
2 years ago
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Helppppp please please please
Firlakuza [10]

Answer:

its d

Step-by-step explanation:

used photo math

4 0
1 year ago
Read 2 more answers
1. Prove or give a counterexample for the following statements: a) If ff: AA → BB is an injective function and bb ∈ BB, then |ff
Fantom [35]

Answer:

a) False. A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1

b) True

c) True

d) B = {1}, A = N, f: N ⇒ {1}, f(x) = 1

Step-by-step explanation:

a) lets use A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1. Here f is injective but 2 is an element of b and |f−¹({b})| = 0., not 1. This statement is False.

b) This is True. If  A were finite, then it can only be bijective with another finite set with equal cardinal, therefore, B should be finite (and with equal cardinal). If A were not finite but countable, then there should exist a bijection g: N ⇒ A, where N is the set of natural numbers. Note that f o g : N ⇒ B is a bijection because it is composition of bijections. This, B should be countable. This statement is True.

c) This is true, if f were surjective, then for every element of B there should exist an element a in A such that f(a) = b. This means that  f−¹({b}) has positive cardinal for each element b from B. since f⁻¹(b) ∩ f⁻¹(b') = ∅ for different elements b and b' (because an element of A cant return two different values with f). Therefore, each element of B can be assigned to a subset of A (f⁻¹(b)), with cardinal at least 1, this means that |B| ≤ |A|, and as a consequence, B is finite.

b) This is false, B = {1} is finite, A = N is infinite, however if f: N ⇒ {1}, f(x) = 1 for any natural number x, then f is surjective despite A not being finite.

4 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!<br><br> Multiply.<br><br> (7x + 3)(4x − 5)
Anni [7]

(7x + 3)(4x -5)

7x*4x+7x(-5)+3*4x+3(-5)

7*4xx-7*5x+3*4x-3*5

7*4xx-7*5x+3*4x-3*5

=28x^2-23x-15

Hope this helps!

Thanks!

-Charlie

3 0
3 years ago
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