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Artist 52 [7]
3 years ago
8

I don’t know where to start someone helpp

Mathematics
2 answers:
tia_tia [17]3 years ago
8 0
20x+10=total dollars
11111nata11111 [884]3 years ago
5 0
Donald will have 20x+10 dollars
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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
3 years ago
23 and 24 I need help please
Gnoma [55]

Answer:

They should purchase the snack with the purple box. This is the answer for 23, i don't know what 24 is sorry.

Step-by-step explanation:

24÷3=8

8×5=40

so the first snacks would cost them $40.

then:

24÷4=6

6×6=36

so the second snacks would cost them $36

5 0
3 years ago
Evaluate (tan 60)(cos 45)
loris [4]

Answer: the tan. Of 60 is .3200

The cos. Of 45 is .5253

Hope this helps :)

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Solve for x. (x - 4)(x - 4) = 0
statuscvo [17]
I believe that x=4 would be the answer
5 0
3 years ago
Read 2 more answers
The length of a rectangle is 6 cm
Tasya [4]

Answer:

12 cm L=6cm W= 2 A=L*W 6*2=12cm

Step-by-step explanation:

7 0
3 years ago
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