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jeyben [28]
3 years ago
6

Find the exact values of the following, giving your answers as fractions.

Mathematics
2 answers:
Stella [2.4K]3 years ago
7 0

Answer:

  • 1/9
  • 1/64
  • 64

Step-by-step explanation:

The applicable rule of exponents is ...

  a^(-b) = 1/(a^b)

Of course, an exponent is used to signify repeated multiplication.

__

3^(-2) = 1/3^2 = 1/(3·3) = 1/9

__

4^(-3) = 1/4^3 = 1/(4·4·4) = 1/64

__

2^6 = 2·2·2·2·2·2 = 64

9966 [12]3 years ago
4 0

Answer:

1/9 1/64 1/64

Step-by-step explanation:

to get full marks on mathswatch.

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2(4+2x)≥5x+5 solve inequality
valkas [14]

Answer:

x ≤ 3

Step-by-step explanation:

2(4+2x)≥5x+5

8+4x≥5x+5

4x-5x≥5-8

-x≥-3

x ≤ 3

6 0
3 years ago
Read 2 more answers
Pam’s Pizza Parlor offers six different toppings for pizza: pepperoni, sausage, meatballs, peppers, onions, and mushrooms. You w
Gwar [14]
I would see 12 different combinations. 
4 0
3 years ago
Choose 5 cards from a full deck of 52 cards with 13values (2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A) and 4 kinds(spade, diamond, h
Delvig [45]

Answer:

a) 182 possible ways.

b) 5148 possible ways.

c) 1378 possible ways.

d) 2899 possible ways.

Step-by-step explanation:

The order in which the cards are chosen is not important, which means that we use the combinations formula to solve this question.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

In this question, we have that:

There are 52 total cards, of which:

13 are spades.

13 are diamonds.

13 are hearts.

13 are clubs.

(a)Two-pairs: Two pairs plus another card of a different value, for example:

2 pairs of 2 from sets os 13.

1 other card, from a set of 26(whichever two cards were not chosen above). So

T = 2C_{13,2} + C_{26,1} = 2*\frac{13!}{2!11!} + \frac{26!}{1!25!} = 182

So 182 possible ways.

(b)Flush: five cards of the same suit but different values, for example:

4 combinations of 5 from a set of 13(can be all spades, all diamonds, and hearts or all clubs). So

T = 4*C_{13,5} = 4*\frac{13!}{5!8!} = 5148

So 5148 possible ways.

(c)Full house: A three of a kind and a pair, for example:

4 combinations of 3 from a set of 13(three of a kind ,c an be all possible kinds).

3 combinations of 2 from a set of 13(the pair, cant be the kind chosen for the trio, so 3 combinations). So

T = 4*C_{13,3} + 3*C_{13.2} = 4*\frac{13!}{3!10!} + 3*\frac{13!}{2!11!} = 1378

So 1378 possible ways.

(d)Four of a kind: Four cards of the same value, for example:

4 combinations of 4 from a set of 13(four of a kind, can be all spades, all diamonds, and hearts or all clubs).

1 from the remaining 39(do not involve the kind chosen above). So

T = 4*C_{13,4} + C_{39,1} = 4*\frac{13!}{4!9!} + \frac{39!}{1!38!} = 2899

So 2899 possible ways.

4 0
3 years ago
Solve for c; a/b+c=d/c
Eduardwww [97]

To solve for c, you need to get c onto one side of the equation, or make it c=__. So what I would do first is subtract a/b from both sides

a/b + c = d/c

-a/b

a/b - a/b + c = d/c - a/b

c = d/c-a/b

8 0
4 years ago
John paid $80 for 5 tickets at an amusement park. Each of the tickets cost the same price. Which equation represent the cost, C,
cupoosta [38]
Since it’s $80 for 5 tickets, we divide 80/5 to get the price of one ticket, giving us

C = 16*n.

Plug in 5 for n and you’ll get 80
3 0
3 years ago
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