BuilderBoi

The prime factorization of 105 is: 5 x 7 x 3

Because there are only 3 prime factors, the 4th digit must be a 1. This makes for \(\color{brown}\boxed4\) possible M's

Distribute the equation, and you get \(20+20i-25i-25i^2\)

After simplifying, you get: \(-25i^2-5i+20\)

Substitute -1 for \(i^2\): \((-25)(-1)-5i+20\)

Simplify again, and we find that the answer is: \(\color{brown}\boxed{-5i+45}\)

1 meter = 1000 millimeters

In his drawing, 1,000 millimeters is 9 millimeters.

Thus, the scale factor is \(\color{brown}\boxed{1000\over9}\)

The formula for the area of a trapezoid is:\({{h(a+b)}\over2} \)

In the equation, h is the height, a is base 1, and b is base 2.

\({{8(9.8+x)}\over2}=59.2\)

\(4(9.8+x)=59.2\)

\(14.8=9.8+x\)

Thus, the base of the other side is \(\color{brown}\boxed{5}\)

I misread the question... DB should be \(4 \sqrt6\), because \(\triangle{CDB}\) is a 30-60-90.

Each roll has a \(1 \over3\) chance of being a square (1 and 4)

Thus, the probability that both rolls are a square is: \({1 \over 3} \times {1 \over 3} = \color{brown}\boxed{1 \over9}\)

The experimental probability is \(13\over25\), because that is the probability-based of the experiment conducted.

However, the theoretical probability is \(1 \over 2\), because theoretically, there are even odds of getting a head or tail. 

Using this, you can find the answer. Feel free to post any questions if needed.

This problem is actually simple because it is a 30-60-90 triangle.

Because we know that \(CD = 8\sqrt3\), \(AD\) must equal \(\sqrt2\times 8\sqrt3 = 8\sqrt6\)

We also know that \(BD\) must equal to \({8\sqrt3}\over\sqrt2\), which can be rewritten as \(4\sqrt6\). 

This means that \(\color{brown}\boxed{AB=12\sqrt6}\)

You can make the equation: \({(x+7.5)}{({2\over3}x+5)}<100\)

The second term is the height of the triangle, and the first term is the base. 

When you solve the equation, you find that \(x\leq4.74\). 

This means that the possible ranges are: \(-7\leq x\leq4\).

Of the values, \(-4\leq x \leq4\) cancel out, making the sum: \(\color{brown}\boxed{-15}\)

CPhill, your answer is incorrect.

The probability that exactly 3 dice show the same is \({5 \choose 3} \times {1 \over 6^3} \times{5 \over6^2} \times 6\). \(5 \choose 3\) ways to pick the rolls that are the same, \({1 \over 6}^3\) chance of 3 successes,  \({5 \over 6}^2\) chance of 2 failures, and 6 different trios of numbers that are the same. 

The probability that exactly 4 dice show the same is \({5 \choose 4} \times {1 \over6}^4\times {5 \over6} \times 6\). With the same logic as above.

The probability that all 5 dice show the same is \({1 \over 6}^5\times6\). 

Add everything up, and you get \(\color{brown}\boxed{23\over108}\)