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PSYCHO15rus [73]
3 years ago
14

What is the domain of the given function? LaTeX: {(3, –2), (6, 1), (–1, 4), (5, 9), (–4, 0)} ( 3 , – 2 ) , ( 6 , 1 ) , ( – 1 , 4

) , ( 5 , 9 ) , ( – 4 , 0 ) Group of answer choices LaTeX: \lbrace x | x = –4, –2, –1, 0, 1, 3, 4, 5, 6, 9 \rbrace { x | x = – 4 , – 2 , – 1 , 0 , 1 , 3 , 4 , 5 , 6 , 9 } LaTeX: \lbrace y | y = –2, 0, 1, 4, 9\rbrace { y | y = – 2 , 0 , 1 , 4 , 9 } LaTeX: \lbrace x | x = –4, –1, 3, 5, 6\rbrace { x | x = – 4 , – 1 , 3 , 5 , 6 } LaTeX: \lbrace y | y = –4, –2, –1, 0, 1, 3, 4, 5, 6, 9 \rbrace
Mathematics
1 answer:
Musya8 [376]3 years ago
3 0

Given:

\text{Function}=\{(3, -2), (6, 1), (-1, 4), (5, 9), (-4, 0)\}

To find:

The domain of the given function.

Solution:

Domain is the set of input values or x-values.

In the given function, the x-coordinates of ordered pairs are 3, 6, -1, 5 and -4. So, domain is the set of these values in ascending order.

The set builder form of domain is

\text{Domain}=\{x|x=-4,-1,3,5,6\}

Therefore, the correct option is C.

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salantis [7]

Answer:

60 nickels

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Step-by-step explanation:

convert $10 to cents

$10 = 1000 pennies

1000 = 5x + 10 (x+10)

1000 = 15x + 100

900 = 15x

x = 60

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check to be sure this adds up to $10

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20 pts please get this right
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Step-by-step explanation:

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Use induction to prove: For every integer n > 1, the number n5 - n is a multiple of 5.
nignag [31]

Answer:

we need to prove : for every integer n>1, the number n^{5}-n is a multiple of 5.

1) check divisibility for n=1, f(1)=(1)^{5}-1=0  (divisible)

2) Assume that f(k) is divisible by 5, f(k)=(k)^{5}-k

3) Induction,

f(k+1)=(k+1)^{5}-(k+1)

=(k^{5}+5k^{4}+10k^{3}+10k^{2}+5k+1)-k-1

=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k

Now, f(k+1)-f(k)

f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-(k^{5}-k)

f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-k^{5}+k

f(k+1)-f(k)=5k^{4}+10k^{3}+10k^{2}+5k

Take out the common factor,

f(k+1)-f(k)=5(k^{4}+2k^{3}+2k^{2}+k)      (divisible by 5)

add both the sides by f(k)

f(k+1)=f(k)+5(k^{4}+2k^{3}+2k^{2}+k)

We have proved that difference between f(k+1) and f(k) is divisible by 5.

so, our assumption in step 2 is correct.

Since f(k) is divisible by 5, then f(k+1) must be divisible by 5 since we are taking the sum of 2 terms that are divisible by 5.

Therefore, for every integer n>1, the number n^{5}-n is a multiple of 5.

3 0
2 years ago
8 x(11-3) + 9 with and exponent of 2
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Answer:

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Step-by-step explanation:

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