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"Other way around" is the same answer. Because here, you can consider absolute value since x and y are nonzero.

what does 3 significant figures mean? nvm i got it

i don't think it is possible

imagine 20x/2=240

x=24

but according to the pythagorean theorem, 20^2=24^2+something squared

you can tell something is wrong

i may be wrong, because i am assuming you mean that the area of abc=240

but if the perimeter of abc is 240, thats alos not possible because two sides of a triangle always has to be greater than the other one.

In radical form, it would be \(\frac{\sqrt{136}}{4}\)

or, \(\frac{2\sqrt{34}}{4}\)or \(\frac{\sqrt{34}}{2}\)

with some logic we can figure out that

\(x=8\)

\(y=2\)

plug it in the equation,

\((8-2)\times(8^2-2^2)\)

\(6\times60\)

\(360\) of course, there's the other way around where x is 2 and y is 8

i'll leave that to you!

9 students\(\times\)3=$27

6 adults\(\times \)7=$42

$27+$42=$69

Here is the really simple way to do it.

No algebra AT ALL!  You dont even have to work out the equation of any of the circles!

The shortest difference between two points is a straight line.

But in this case a part of the problem is that a 4 foot vertical drop must be included.

I have included it at the start but it can be included at the end if you want. I called the end point of it A'

Now use pythagoras to find distance A'D     5,12,13 is a pythagorean triad so  A'D=13

The rope is 7 foot

13 - 7 = 6 feet left!  

Easy peazy!

I personally got \(4+\sqrt{5}\) can someone double check?

What asinus has said (sort of)  is that a part of the vertical drop is the 4feet.

Which on my number plane is the point (0,5)

THEN

the the shortest path to D is a straight line.  The rope will cover the first 7 feet.  then you have to work out what is left.

That is much easier than how I did it.  

EXCELLENT thinking asinus!

z² +az + b = 0, roots are p and q

p + q = -a, pq = b

(-7 + 3i) + (-7 - 3i) = -14 = -a => a = 14

(-7 + 3i)(-7 - 3i) = 49 - 9(-1) = 58 = b => b = 58

z² + 14z + 58 = 0 => \(x = {-14 \pm \sqrt{14^2-4(1)(58)} \over 2(1)} \)

\(x = {-14 \pm \sqrt{-36} \over 2} \)

z = (-7 + 3i) or (-7 - 3i)

x = 5 - sqrt(69), y = 5 + sqrt(69)

x = 5 + sqrt(69), y = 5 - sqrt(69)

sheesh

There are three kinds of planes: the six cube faces, the six that pass through 4 points (opposite edges), and the eight that pass through only 3 vertices, for a total of 6 + 6 + 8 = 20.

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 \(\color{brown} 288\).

These angles are 30, 60, and 90, which has a side ratio of 1, sqrt3, 2

Since the shortest side is 12, the longest side is hence 24

Thanks asinus,

I had not actually understood what you had said before but now this comment totally explaines it!

Yes that would be a good idea!

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 \(\color {red}567\)

What is the largest possible length of the shortest side of the parallelogram?

Hello Guest!

The longest side of the parallelogram is when the diagonals are perpendicular to each other. It will be a four-sided rhombus. The length of the sides is

\(s=\sqrt{(\frac{10}{2})^2+(\frac{6}{2})^2}\\ \color{blue}s=5.831\)

  !

Hi Melody!

Graphic: I would have extended the DC line to the center of the lower circle. So she is the image of my thoughts on the solution.

  !

There are:  (8, 144, 2584, 46368, 832040, 14930352, 267914296, 4807526976, 86267571272, 1548008755920, 27777890035288, 498454011879264, 8944394323791464, 160500643816367088, 2880067194370816120, 51680708854858323072, 927372692193078999176) >>Total = 17 terms are divisible by 8

There should be: 777 zeros in base 10.

because you guys had different answers, I did it myself and I found idk ( I didn't count) numbers of which 24 numbers are distinct.

1138, NO
1183, NO
1146, NO
1164, NO

1226, NO
1262, NO 
1234, --------------------------- 1
1243, --------------------------- 2

1318, NO
1381, NO
1324, ---------------------------- 3
1342, ----------------------------- 4

1416, NO
1461, NO
1423, ------------------------------ 5
1432, ------------------------------ 6

1614, NO
1641, NO
1622, NO

1813, NO
1831, NO

2126, NO
2162, NO
2134, ----------------------------- 7
2143, ----------------------------- 8

2216, NO
2261, NO
2223, NO
2232, NO

2314, -------------------------------- 9
2341, -------------------------------- 10
2322, NO

2413, -------------------------------- 11
2431, -------------------------------- 12

2612, NO
2621, NO 

3118, NO
3181, NO
3124, -------------------------------- 13
3142, --------------------------------- 14

3214, -------------------------------- 15
3241, -------------------------------- 16
3222, NO

3412, -------------------------------- 17
3421, --------------------------------- 18

3811, NO

4116, NO
4161, NO
4123, ---------------------------------- 19
4132, ------------------------------------ 20

4213, ------------------------------------- 21
4231, ------------------------------------- 22

4312, ------------------------------------ 23
4321, ------------------------------------ 24

4611, NO

6114, NO
6141, NO
6122, NO

6212, NO
6221, NO

6411, NO

8113, NO
8131, NO

8311, NO

This was just like a check that geno's answer was correct... .

thus there are 24 ways.