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3 years ago

Select the sets that are not functions.

Mathematics

2 answers:

Anarel [89]3 years ago

7 0

A cuz of how its functions
and D cuz of how it uses the numbers i guess

egoroff_w [7]3 years ago

5 0

Answer with Step-by-step explanation:

A function is a map from a set A to another set B such that

the image of any point in a has a unique image in set B.

A = {(1, 2), (2, 3), (3, 4), (4, 5)}

It is a function since image of 1,2,3 and 4 is unique.

B = {(1, 2), (2, 1), (3, 0), (4, -1)}

It is a function since image of 1,2,3 and 4 is unique.

C = {(-1, 3), (0, 3), (1, 3), (2, 3)}

It is a function since image of -1,0,1 and 2 is unique.

D = {(1, 1), (1, 2), (1, 3), (1, 4)}

It is not a function since image of 1 is not unique.

E = {(6, 2), (7, 3), (6, -1), (5, 4)}

It is not a function since image of 6 is not unique.

Hence, sets that are not functions are:

D = {(1, 1), (1, 2), (1, 3), (1, 4)}

E = {(6, 2), (7, 3), (6, -1), (5, 4)}

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1) We can get that number hit and trial method.

It is said that "two three-digit numbers and wrote one six-digit number, which is equal to seven times their product".

Let us check number 143.

If we write 143 two times without any sign in between , we get 143143 ( a six digit number).

But if we multiply 143 × 143 , we get 20449.

7 times of 20449 equals 143143.

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2) Let us assume required prime numbers are a, b and c.

Product of a, b and c= abc.

Sum of a, b and c = a+b+c.

"Their product is five times greater than their sum."

Therefore,

abc = 5(a+b+c) ----------------------equation (1)

Now, let us take first prime numbers 2 and second 5.

Plugging a=2 and b=5.

2×5×c = 5(2+5+c).

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10c = 35 +5c.

Subtracting 5c from both sides, we get

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Let us check by taking a=2 and b=7.

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Let us take a=2 and b=11, we get

2×11×c = 11(2+11+c).

c=13 (A prime)

If we take a=2 and b=17, we get

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c=19 (A prime).

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Answer:

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Step-by-step explanation:

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nadezda [96]

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