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Iteru [2.4K]
3 years ago
15

What are the solutions to the equation?

Mathematics
1 answer:
mario62 [17]3 years ago
7 0

Answer:

X=-4 and x=10

Step-by-step explanation:

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At soccer practice on Saturday morning, Winston's team practiced dribbling for 25 minutes and practiced shooting for 15 minutes.
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10h20m AM

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PLEASE HELP FAST The Booster Club raised $30,000 for a sports fund. No more money will be placed into the fund. Each year the fu
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c. $24,435

Step-by-step explanation:

30000(1 - .05)^{4} = 24,435.19

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It takes Dariya 35 seconds to download 5 songs from the Internet. How can the number of seconds it would take Dariya to download
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Step-by-step explanation:

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Write an expression to represent the length of a string in yards in terms of the string's length, x, in inches
Sergio039 [100]

Answer:

y = \frac{x}{36}

Step-by-step explanation:

Given

Length of string = x inches

Required

Determine the length in yards

Represent the equivalent length with y

Such that

y = x\ inches

Divide x by 36 to get equivalent in yards

y = \frac{1}{36} * x

y = \frac{x}{36}

Hence, the equivalent in yards is \frac{x}{36}

3 0
3 years ago
Suppose that five ones and four zeros are arranged around a circle. Between any two equal bits you insert a 0 and between any tw
PolarNik [594]

Answer:

Using <u>backward reasoning</u> we want to show that <em>"We can never get nine 0's"</em>.

Step-by-step explanation:

Basically in order to create nine 0's, the previous step had to have all 0's or all 1's. There is no other way possible, because between any two equal bits you insert a 0.

If we consider two cases for the second-to-last step:

<u>There were 9 </u><u>0's</u><u>:</u>

We obtain nine 0's if all bits in the previous step were the same, thus all bit were 0's or all bits were 1's. If the previous step contained all 0's, then we have the same case as the current iteration step. Since initially the circle did not contain only 0's, the circle had to contain something else than only 0's at some point and thus there exists a point where the circle contained only 1's.

<u>There were 9 </u><u>1's</u><u>:</u>

A circle contains only 1's, if every pair of the consecutive nine digits is different. However this is impossible, because there are five 1's and four 0's (we have an odd number of bits!), thus if the 1's and 0's alternate, then we obtain that 1's that will be next to each other (which would result in a 1 in the next step). Thus, we obtained a contradiction and thus assumption that the circle contains nine 0's after iteratins the procedure is false. This then means that you can never get nine 0's.

To summarize, in order to create nine 0's, the previous step had to have all 0's or al 1's. As we didn't start the arrange with all 0's, the only way is having all 1's, but having all 1's will not be possible in our case since we have an odd number of bits.

<u />

5 0
3 years ago
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