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solniwko [45]
3 years ago
9

HELP !!!! Equations with 1 no or infinite solutions

Mathematics
2 answers:
ryzh [129]3 years ago
6 0

1. v=0

2. No solution

3. m=0

4. No solution

5. p= -7

6. x=5

7. -12 = - 12

8. x=5

9. 3=3

10. -32= -32

11. x=0

12. No solution

I'm sorry if anything is wrong.

kari74 [83]3 years ago
4 0

Step-by-step explanation:

Ok Hold UP

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I put the wrong picture on the last one but this is still due soon
fgiga [73]

Answer:

The point ”A“ is the point that corresponds to the last entry in the table.

Step-by-step explanation:

5 0
3 years ago
Pamela has 4 pink ribbons, 3 green ribbons, and 1 blue ribbon. what fraction of Pamela's ribbons are green?
Brrunno [24]
The total number of ribbons is 8 because 4 + 3 + 1 is 8
The number of green ribbons is 3
That means that 3 out of the 8 ribbons are green
Answer: 3/8
6 0
3 years ago
Read 2 more answers
Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.
frutty [35]
<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

2^n\leq 2^{n+1}-2^{n-1}-1

where n is a positive integer.

  • Let us take n=1

then we have:

2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2

Hence, the result is true for n=1.

  • Let us assume that the result is true for n=k

i.e.

2^k\leq 2^{k+1}-2^{k-1}-1

  • Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u>  2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Let us take n=k+1

Hence, we have:

2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)

( Since, the result was true for n=k )

Hence, we have:

2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2

Also, we know that:

-2

(

Since, for n=k+1 being a positive integer we have:

2^{(k+1)+1}-2^{(k+1)-1}>0  )

Hence, we have finally,

2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

2^n\leq 2^{n+1}-2^{n-1}-1  where n is a positive integer.

6 0
3 years ago
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Fantom [35]

Answer:

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Step-by-step explanation:

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6= k × 2/5

K = 6÷2/5

K= 6×5/2

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5 0
3 years ago
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Are the dotted lines in the shape parallel, perpendicular or intersecting?
slava [35]

Answer:

<em>Parallel</em>

Step-by-step explanation:

The dotted lines are parallel because they have the same <u>gradient</u> to each other, this means the two dotted lines will <u>never</u> touch each other.

Have a great day <3

8 0
3 years ago
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