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Hitman42 [59]
3 years ago
9

N(U)=90, n(A)=35, n(B)=45 Then n(A’ u B)= ?

Mathematics
1 answer:
Akimi4 [234]3 years ago
6 0
I did y’all win the first time and it is a great way home game tmr is the best season for ya
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What is the value of x?
DENIUS [597]

2x+x+47

3x+47

3x+47=x+93

2x+47=93

2x=46

x=46/2

x=23

You must use all the equations that makes up the x. Using all the equations will cover the area than just using part of it.

4 0
3 years ago
Read 2 more answers
Jeanne makes hair bows to sell at the craft fair. Each bow requires 1.5 yards of ribbon.
UkoKoshka [18]

Answer:

a) 378 foot

b) 3780¢

c)10^3\times 2^1\times 3^3\times 7^1

Step-by-step explanation:

Given: Ribbon required to make 1 bow = 1.5 yards

We know that 1 yard = 3 foot

a) Number of bows Jeanne wants to make = 84

Foot of ribbon required to make 84 bows

= Number of bows Jeanne wants to make × Ribbon required to male 1 bow

= 84 × 1.5 yards

= 126 × 3 foot = 378 foot

Thus, Jeanne requires 378 foot of ribbon to make 84 bows.

b) Cost of ribbon = 10¢ per foot

Total money required to buy ribbon

= Cost of ribbon × Amount of ribbon required in foot

= 10 × 378

= 3780.00¢

c) Number of bows manufacture wants to make = 84 × 1000

= 84,000 bows

Amount of ribbon required to make bows

= 1000 × Amount of ribbon required to make 84 bows

=  1,000 × 378 = 378000 foot

To write it in exponential form:

378000 = 10^3\times 2^1\times 3^3\times 7^1

8 0
3 years ago
Can you help me with this question?
goldenfox [79]
We can start by saying that 4/3w+ 3/3w = 133.
we can simplify to 7/3w = 133.
multiply both sides by 3 to get 7w = 399.
divide both sides by 7 to get 57.
Mr. Wong's mass is 57 kg.
3 0
3 years ago
Em um colégio,de 50 alunos ,40 gostam de sorvete de chocolate,35 gostam de sorvete de creme e 30 gostam dos dois sabores .Quanto
GarryVolchara [31]

Answer:

i do not speak ur language

Step-by-step explanation:

5 0
2 years ago
Find the number of permutations of the first 9 letters of the alphabet, taking 4 letters at a time.
Vinil7 [7]

Solution-

First 9 letters of alphabet are- A,B,C,D,E,F,G,H,I

Total number of ways of selection of 4 letters from 9 alphabets = 9C4

=9!÷((4!).(9-4)!) = 9!÷(4!×5!) = (9×8×7×6)÷(24) = 126

The number of ways of arranging these 4 numbers = 4! = 24

∴ Total number of possible permutations = 9C4×4! = 126×24 = 3024

∴ option number 2 is correct.

8 0
3 years ago
Read 2 more answers
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