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Tcecarenko [31]
3 years ago
8

pLEASE HELP first how do I get to the next question and what am I supposed to graph and what’s the answer

Mathematics
2 answers:
marta [7]3 years ago
8 0

Answer:

Option D will be your answer

▪ no solution

Step-by-step explanation:

hope it's helpful...

Zielflug [23.3K]3 years ago
6 0
No solution..... so last one
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Greastest common factor please help me
Hatshy [7]

Answer:

c

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Marc-Andre Fleury made saves on 27 of the first 29 shots in the hockey game Find experimental probability that he will make a sa
JulsSmile [24]
Experimental probability is number of observed specific outcomes divided by the total number of observations...in this case the experimental probability of making a save is:

27/29≈0.931
8 0
3 years ago
Factor completely.
Serga [27]
Factor the polynomial:

4u² – 20u + 25

Rewrite – 20u as – 10u – 10u, and then factor it by grouping:

= 4u² – 10u – 10u + 25

= 2u * (2u – 5) – 5 * (2u – 5)

= (2u – 5) * (2u – 5)

= (2u – 5)² <––– this is the answer.

I hope this helps. =)
5 0
3 years ago
Please help me. i want to have this finished.
Anuta_ua [19.1K]
We know that y is equal to x-2. We can just substitute x-2 for y since y is equal to it. 

10x- 9(x-2)=24

Distribute.

10x- 9x+18= 24

x+18= 24

Subtract 18 on both sides.

x= 6

Now, plug in x.

y= (6)-2

y= 4

We can check this to see if this works:

4= 6-2, 4=4 

10(6)- 9(4)= 24

60-36= 24, 24=24

x=6 and y=4

I hope this helps!
<em>~kaikers</em>
5 0
3 years ago
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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