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Sergio039 [100]
3 years ago
6

Factor the polynomials completely. If the polynomial cannot be factored, write "prime."

Mathematics
1 answer:
Monica [59]3 years ago
7 0

Answer:

4) -(m+12)(m-12)

5) (y+12)^2

6) (x+5)(x-5)(x^2-25)

7) (a+9)(-y+7)

8) Prime

9) Prime

Hope this helps!


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Please help me solve this..​
Nataliya [291]

Answer:

5

Step-by-step explanation:

Let a = number of pieces of chocolate bought by Amin.

Let b = number of pieces of chocolate bought by Bob.

b = 2a

(a - 3)(b - 3) = 14

ab - 3a - 3b + 9 = 14

a(2a) - 3a - 3(2a) = 5

2a^2 - 3a - 6a = 5

2a^2 - 9a - 5 = 0

(2a + 1)(a - 5) = 0

2a + 1 = 0  or  a - 5 = 0

a = -1/2  or  a = 5

Amin cannot have bought -1/2 pieces of chocolate, so we discard the soluion a = -1/2.

a = 5

Answer: 5

3 0
2 years ago
Read 2 more answers
Will someone please answer this ASAP??
nirvana33 [79]
SA = 2bA (base area) + LA (lateral area)

bA = r^2*pi = 13^2pi = 169pi cm^2

LA = 2r*pi*h = 2*13*pi*88,4 = 2298,4pi cm^2

SA = (2*169pi)+2298,4 pi = 338pi+2298,4pi = 2636,4pi
3 0
3 years ago
Bob has a standard deck of playing cards. If he randomly draws a J, K, 2, and 2, what is the probability that the next card he d
nlexa [21]

Answer:

The correct answer has already been given (twice). I'd like to present two solutions that expand on (and explain more completely) the reasoning of the ones already given.


One is using the hypergeometric distribution, which is meant exactly for the type of problem you describe (sampling without replacement):


P(X=k)=(Kk)(N−Kn−k)(Nn)


where N is the total number of cards in the deck, K is the total number of ace cards in the deck, k is the number of ace cards you intend to select, and n is the number of cards overall that you intend to select.


P(X=2)=(42)(480)(522)


P(X=2)=61326=1221


In essence, this would give you the number of possible combinations of drawing two of the four ace cards in the deck (6, already enumerated by Ravish) over the number of possible combinations of drawing any two cards out of the 52 in the deck (1326). This is the way Ravish chose to solve the problem.


Another way is using simple probabilities and combinations:


P(X=2)=(4C1∗152)∗(3C1∗151)


P(X=2)=452∗351=1221


The chance of picking an ace for the first time (same as the chance of picking any card for the first time) is 1/52, multiplied by the number of ways you can pick one of the four aces in the deck, 4C1. This probability is multiplied by the probability of picking a card for the second time (1/51) times the number of ways to get one of the three remaining aces (3C1). This is the way Larry chose to solve the this.

Step-by-step explanation:


6 0
3 years ago
Read 2 more answers
. The diameter of Bubba's bubble is
Oksana_A [137]

Answer:

10.22

Step-by-step explanation:

2.45 * 10 =20.45 - 3.67 * 10 =30.67

30.67 - 20.45

30 - 20 = 10

.67 - .45 = .22

30.67 - 20.45 = 10.22

8 0
2 years ago
Please help me with precal
Doss [256]

Consider the function f(x)=\sqrt{x}. This function has:

  • the domain x\in [0,\infty);
  • the range y\in [0,\infty).

If the domain of unknown function is [a,\infty), then x\ge a or x-a\ge 0. This means that you have \sqrt{x-a} as a part of needed function.

If the range of unknown function is [b,\infty), then y\ge b. This means that you have to translate function b units up and then the expression of the function is

y=\sqrt{x-a}+b.

Answer: correct choice is B.

6 0
3 years ago
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