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ivann1987 [24]
3 years ago
12

Which equation has no solution?

Mathematics
1 answer:
san4es73 [151]3 years ago
4 0

Answer:

c

Step-by-step explanation:

hope this helps

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Find the perimeter of the square<br> With one side 2 1/2 yards
Lilit [14]

Answer:

12

Step-by-step explanation:

2.5*4=12

if each side of a square is the same length then yes my answer is correct

4 0
3 years ago
What's the area of this figure?
Westkost [7]
Its a triangle height = 12-3 = 9 and length = 4+5 = 9      9*9 = 81
and a rectangle height = 3  and length = 5     3*5 = 15
15+81 = 96
6 0
3 years ago
A company's board of directors wants to form a committee of 2 of its members. There are 8 members to choose from. How many diffe
irga5000 [103]

Answer:

105 different committees can be formed.

6 0
3 years ago
A square is changed into a rectangle by increasing the length of the square by 5 units to be (x + 5) and increasing the width by
gladu [14]
Area of the rectangle = x^2 + 8x + 15 square units

Explanation:

Length = x + 5
Breadth = x + 3

Area = l x b = (x + 5)(x + 3) = x(x + 3) + 5(x + 3) = x^2 + 3x + 5x + 15 = x^2 + 8x + 15 square units
4 0
3 years ago
You are given a black box f : Z10 → Z10 that contains either a random permutation or a random function. Your distinguisher is al
SashulF [63]

Solution :

The function :    $f: Z_{10} \rightarrow Z_{10}$   be a random permutation.

f is a permutation on $Z_{10}$ ,  i.e. f is permutation 10.

Now we know that the total number of distinct permutation on to symbolize 10!.

Each of these 10! permutation to a permutation function  $f: Z_{10} \rightarrow Z_{10}$  

Therefore, total number of permutation functions  $f: Z_{10} \rightarrow Z_{10}$   are 10!.

Now we want the total number of permutation functioning :

$f: Z_{10} \rightarrow Z_{10}$   such that  f(0) = 0 and f(1)= 1

Now we notice that when f(0)=0 and f(1)=1, then two symbol '0' and '1 are fixed under permutation f.

So essentially when f(0) = 0 and f(1) = 1, f becomes permutation on 8 symbol.

Total number of permutation functioning  $f: Z_{10} \rightarrow Z_{10}$  , f(0)=0 and f(1)=1 are 8!

Now we want the probability that a random permutation  $f: Z_{10} \rightarrow Z_{10}$   satisfies f(0) = 0 and f(1) = 1.

The number of permutation function  $f: Z_{10} \rightarrow Z_{10}$  , i.e.

The probability that a random permutation  $f: Z_{10} \rightarrow Z_{10}$   satisfies f(0) = 0 and f(1) = 1 is

$\frac{8!}{10!} = \frac{8!}{10 \times 9\times 8!} =\frac{1}{10 \times 9}=\frac{1}{90}$

Therefore, the probability that a random permutation  $f: Z_{10} \rightarrow Z_{10}$   satisfies f(0)= 0 and f(1)=1 is  $\frac{1}{90}$

6 0
3 years ago
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