BuilderBoi

https://web2.0calc.com/questions/algebra-question_120#r1. I answered the question here

For the expression to be undefined, it has to be \(x \over 0\). The only ways to get \(a^3-a\) equal to \(0 \) is when \(a \) is \(\color{brown} -1, 0, 1 \)

First, we need to simplify \(4(x + 7)(2 - x) + 4(x + 7)(8 - x)\)into \(−8x^2−16x+280\)

The maximum value of any quadratic equation is \(c - b^2/4a\), when the quadratic is in \(ax^2 + bx + c\). By subsituting, we find that the maximum value for this equation is \(\color{brown} 288\).

Since the series is geometric, you have to find the ratio between \(2\) and \(1 \over 5 \). To get from \(2\) to \(1 \over 5\), you multiply by \(1 \over 10\). This means that the \(5\)th term is \({2 \times {1\over10}^4} = {\color {blue} {1\over 5000}}\)

\(\Delta OCD\) is an isosceles triangle, with \(OC\) and \(OD\) both equal to \(5\).

Label the midpoint of \(CD\) as \(B\). Draw a line connecting \(O\) to \(B\). You have a 60-30-90 triangle, where the hypotenuse is \(5\). This means that \(OB\) is \(2.5\). Using the Pythagorean theorem, you find that \(CB\) is \(\sqrt{18.75}\), which means that \(CD\) is \(2\sqrt{18.75}\), which can be simplified to \(\color{purple} 5\sqrt3\).

Solution 1:

The slope between \((5,5)\) and \((3,1)\) is \({{1-5} \over {3-5} } = 2\) 

To get from x coordinate \(3\) to x coordinate \(-6 \), you need to subtract \(9\), meaning that the y-value would be \(18\) less than \(1\), s \(c = \color {red} -17 \).

Solution 2:

Using the method above, you know that the slope is -2, so when you plug it into the equation \(y=mx+b\), with \(x = 5\), \(y = 5\), and \(m= 5 \) (coordinates of 1 point doesn't matter which, and slope respectively), you get: \(5 = 2 \times 5 + b\), where \(b = -5\), giving us the equation \(y=2x-5\).

Plug in \(x = -6\), and you find that \(c = \color{Red} 17\)

Let's count the multiples of 21 that are less than 100: 21, 42, 63, 84

 The next possible range is \(>550\), (because anything less always has a digit less than 5 in the hundreds or tens place):

The multiples are: 567, 588, 609...

We can see that the smallest number with no digit less than 5 that is divisible by 21 is \(\color {red}567\)

There is 5 options of where to put the A. There is 4 options of where to put the D, because the A takes up one spot. FOr the remaining 3 slots, there is \(3^3 = 27 \) ways to arange them, so the answer is \(27 \times 20 = \color{red}540 \) ways.