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Let the cost of the bench ==B

The cost of the garden table ==B + 42

B + B + 42 ==742

2B  +  42 == 742

2B == 742 - 42 ==700

B == 700 / 2 ==350 - cost of the bench

350  +  42 ==392 - cost of the garden table.

2-digit multiple ==18

3-digit multiple ==108

4-digit multiple ==1008

The sum ==18 + 108 + 1008 ==1,134

3x=22-7

3x=15

x=5

5 times 2=10

2(5)=10

2x=10

7.25 1/4 is equivalent to 0.25

(4, 6, 6) , (5, 5, 6) , (5, 6, 5) , (6, 4, 6) , (6, 5, 5) , (6, 6, 4) , Total =  6 such combinations.

6 / 6^3 ==6 / 216 ==1 / 36

Two-digit multiple: 18

Three-digit multiple: 117

Four-digit multiples: 1116

Answer: 18 + 117 + 1116 = 1251

The answer is 80234.

Here is my attempt:

MATHCOUNTS ==10 Letters with 2 "Ts" gives us 9 distinct letters:

Drawing 4 letters at random, gives the following number of permutations:

Since the letters are replaced (or repeats are allowed), then we have:

9 x 9 x 9 x 9 ==9^4 ==6,561 permutations

Since the letters: N U T S can appear in 4! ==24 permutations, then:

The probability is: 24 / 6,561 ==8 / 2187==0.36579789 67 %

Note: somebody should check this!!

(1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 6, 7, 8, 8, 8, 9, 9, 10, 10, 10, 11, 12, 12, 12, 12, 12, 13, 14, 14, 15, 15, 15, 16, 16, 16, 17, 18, 18, 18, 18, 19, 20, 20, 20, 20, 21, 22, 24, 24, 24, 24, 25, 26, 27, 28, 28, 30, 30, 30, 30, 32, 32, 33, 34, 35, 36, 36, 36, 36, 38, 39, 40, 40, 40, 42, 42, 44, 45, 45, 48, 48, 48, 50, 51, 52, 54, 54, 55, 56, 57, 60, 60, 60, 60, 64, 65, 66, 68, 70, 72, 72, 75, 76, 78, 80, 80, 84, 85, 90, 90, 95, 96, 100, 102, 108, 114, 120)>>Total = 120 possible products

(5, 5, 10, 10, 10, 15, 15, 15, 20, 20, 20, 20, 25, 30, 30, 30, 30, 35, 40, 40, 40, 45, 45, 50, 55, 60, 60, 60, 60, 65, 70, 75, 80, 80, 85, 90, 90, 95, 100, 120) >>Total = 40 products that are multiples of 5

Therefore, the probability is: 40 / 120 ==1 / 3

The Formula for the area of a trapezoid is \(A= \frac{a+b}{2}h\)  so plugging in the corresponding numbers, the area of the trapezoid is 81

Ok so it's exact/simplified form is \(180 \sqrt{3}\)

To answer this question, you need to get each number to the same root. You cannot do anything with them now, but by giving them a square root in common you can simplify this expression.

There are 34 integers x.

p(sqrt(3)) = 12

There was 4.45 kg in Bag A at first.

This is A O P S homework.  Do not cheat.

Thanks :)

I didn't notice that 343 was a perfect cube lol 

if we call the capacity of Marshfield Stadium = x, we have

Bowden Stadium =x+4127

x+4127+x=210355

2x=206228

x=103114

Marshfield Stadium Capacity : 103114

Bowden Stadium Capacity : 107241

:]

Thanks proyaop for finding that answer.

This was a great question and the answer by Oofrence was really good.

AutumStorm:

Please note that your number can be expressed as the sum of 2 perfect cubes. That is:

1000343 = 100^3 + 7^3  and  100 + 7 =107. This 107 is ALWAYS a divisor of 1000343. That is, 107 divides  

1000343 evenly with no remainder. Then, the only thing you have to determine is to find out whether 107 is a prime number or not. If you don't know whether it is a prime or not, you may try the sieve method of trying to divide it by all prime numbers from 2 to sqrt(107) ==<11, which is only 4 such primes (2, 3, 5, 7). You will quickly see that it is not divisible by any of the 4 primes. Therefore, it must be a prime number. Good Luck.

\(12w - 27 = -34 + 14w \)

\(-27 = -34 + 2w \)

\(7 = 2w\)

\(w = 7/2\)

or

\(w=3.5\)

:]

Thanks :) Basically, I need to see that 343 is a perfect cube. 

Not me feeling kinda dumb rn lol

Also, guest, please stop complaining about people's answers if you're not answering yourself. (reply number one) 

:) :)