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raketka [301]
3 years ago
14

A high school basketball team has a budget specifically for towels and extra basketballs. A towel costs $3 and a basketball cost

s $18. Assume the budget is $450 and that 30 towels are needed for the team for the entire season. Use a standard form equation to determine how many basketballs may be purchased by the team.
Mathematics
1 answer:
bonufazy [111]3 years ago
4 0
An equation that can be used is 3T + 18B = 450. Since they know they need 30 towels, substitute 30 for T. The equation becomes 3(30) + 18B = 450. Simplifying, 90 + 18B = 450. Subtract 90 from each side, 18B = 360. Finally, divide each side by 18, B = 20. The maximum number of basketballs that may be purchased is 20.
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Find M of angle ABC
pochemuha

Answer:

25

Step-by-step explanation:

\sf Answer \begin{cases}\sf \rightarrowtail \:( 3x - 6) + (2x + 25) + (x + 11) = 180 \\  \sf \rightarrowtail \: 3x - 6 + 2x + 25 + x + 11 = 180 \\  \sf \rightarrowtail \: 3x + 2x + x - 6 + 25 + 11 = 180 \\  \sf \rightarrowtail \: 6x + 30 = 180 \\  \sf \rightarrowtail \: 6x = 180 - 30 \\  \sf \rightarrowtail \: 6x = 150 \\  \sf \rightarrowtail \: x =  \frac{150}{6}  \\  \sf \rightarrowtail \: x = 25\end{cases}

5 0
3 years ago
for each babysitting job anna charges 2.50 for bus fare plus $8 per hours she worked. She charged 3.50 for her last babysitting
julsineya [31]
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5 0
3 years ago
Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$
Viktor [21]

When n=0, we have

5^{3^0} + 1 = 5^1 + 1 = 6

3^{0 + 1} = 3^1 = 3

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for n=k, that

3^{k + 1} \mid 5^{3^k} + 1

Now for n=k+1, we have

5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)

so we know the left side is at least divisible by 3^{k+1} by our assumption.

It remains to show that

3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1

which is easily done with Fermat's little theorem. It says

a^p \equiv a \pmod p

where p is prime and a is any integer. Then for any positive integer x,

5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3

Furthermore,

5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3

which goes all the way down to

5^{3^k} \equiv 5 \pmod 3

So, we find that

\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3

QED

5 0
1 year ago
Can sumone plzhelp..Apply the distributive property to factor out the greatest common factor of all three terms .
Nat2105 [25]

Answer:

3(3a - 6b + 7c )

Step-by-step explanation:

The greatest common factor of 9, 18 and 21 is 3

Factor out 3 from each term

9a - 18b + 21c

= 3(3a - 6b + 7c) ← in factored form

6 0
3 years ago
Read 2 more answers
What is 9 29/40 as a decimal
earnstyle [38]

Answer:

The answer is 9.725

Step-by-step explanation:

We have been given the mixed fraction 9\frac{29}{40}

First we will concert this to improper fraction.

\frac{40(9)+29}{40}

= \frac{389}{40}

Now we will divide this using long division or calculator and get the answer  9.725.

5 0
3 years ago
Read 2 more answers
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