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what does $[ABC] = 240$ represent? 

Hello Guest!

The greatest possible length of the shortest side of the parallelogram is greater than two by the smallest possible length.

  !

here is a representative image i made:

since both diagonals cot eachother in the middle, they form 4 triangles; by looking at the image, we can say that:

$ \overline{AO}=\frac{6}{2} \ \therefore \ \overline{CO}=\frac{6}{2}  $

$ \overline{DO}=\frac{10}{2} \ \therefore \ \overline{BO}=\frac{10}{2}  $

now there are two little rules for triangles 

1) sum of any two sides $>$ third side.

2) difference of any two sides $<$ third side. ( commonly used only to prove whether  the triangle exists or not )

lets consider the $\triangle BOC $ where side $\overline{BC}=$third side (the third and largest side of the triangle, but the smallest side of the parallelogram)

1)  $ \frac{10}{2}+\frac{6}{2} > \text{ third side} $

$ 5+3> \text{ third side} $

$ 8>  \text{ third side}  $

changing the sides, would also change the 'direction' of the less-than-sign, so logically, it would become larger-than-sign:

$   \text{ third side} <8 $

To start, note that $\frac{x}{x^{2/3}}=x^{1-\frac{2}{3}}=x^{1/3}$. Also note that we can rewrite the cube root with a fractional exponent, so $\sqrt[3]{x}=x^{1/3}$. Using these pieces of information, rewrite the given equation as:$$4x^{1/3}-2x^{1/3}=7+x^{1/3}$$Move all $x^{1/3}$ terms to one side and when simplified, you get x=343

yw

Slope, m, between the first two points     (y1-y2)/(x1-x2) = 4/2 = 2  

    slope between the next two points is also = 2       so      (1-c)/(3--6) = 2

                                                                                               1-c = 18

                                                                                                c = -17

From a1 to a2   is    a1 * r     so r = 1/10

   starting form second term   1/5 * r * r * r   = a5 = 1/5 * 1/1000 = 1/5000

so we have  $ \{  \underbrace{2}_\text{ first term },\underbrace{\frac{1}{5}}_\text{ 2nd term } , \underbrace{...}_\text{ 2 other terms },  \underbrace{a_5}_\text{ 5th term } \}  $

the formula to find any term of this geosequence is

$a_n=ar^{n-1}$ where $a_n=n^{th} term$    ;   $ a=\text{first term}$   ;   $r=\text{common ratio}$

so far we have $ a_5=2r^{5-1}   $

to find the ratio we just write this simple equation $  2\cdot \:r=\frac{1}{5}  \implies  r=\frac{1}{10} $

now that we know what the common ratio is, going back and plugging it in the formula we get:

$ a_5=2\left(\frac{1}{10}\right)^{5-1}  $

$   a_5=2\cdot \frac{1^4}{10^4}  $ 

$ a_5=2\cdot \frac{1}{10^4}  $

$   a_5=\frac{2}{10^4}  $

$ \boxed{a_5=\frac{2}{10000}}  \Leftrightarrow \boxed{ a_5=\frac{1}{5000}}  $

first of all lets evaluate a bit:

$ \frac{a+3}{a^3-a} $

$ \frac{a+3}{a\left(a^2-1\right) } $

$ \frac{a+3}{a\left(a+1\right)\left(a-1\right)}  $

now, any rational expression will be undefined when the denominator is $=\: 0$ right? because dividing by 0 ( consider it like like something which does not exist ) just makes no sense. so we note that when $a\left(a+1\right)\left(a-1\right)=0$, that expression is undefined:

$a\left(a+1\right)\left(a-1\right)=0$

$ \left(a+1\right)\left(a-1\right)=\frac{0}{a}  $

$  \left(a+1\right)\left(a-1\right)=0   $       $ \implies \begin{cases} a+1=0  \\ \ \ \ \ \text{and}  \\ a-1=0   \end{cases} \implies \begin{cases} a=-1  \\ \ \ \text{and}  \\ a=1   \end{cases}   $  

thus, whenever $\boxed{a=1,\: -1}$ the fraction becomes undefined

according to this website, https://www.mathwarehouse.com/answered-questions/multiples/what-are-the-multiples-of-21-solved.php all the multiples of 21 are:

$ 21, 42, 63, 84, 105, 126, 147, 168, 189, 210, 231, 252, 273, 294, 315, 336, 357, 378, 399, 420, 441, 462, 483, 504, 525, 546, 567, 588, 609, 630, 651, 672, 693, 714, 735, 756, 777, 798, 819, 840, 861, 882, 903, 924, 945, 966, 987, 1008, 1029 $

we can right away cancel all the ones which are $< \: 500$ right?

its just easy from here -- looking from the left to the right we have to find which number has no figures $< \: 500$ :

$  \color{black}{504, 525, 546,} \color{royalblue}{567}   \color{black}{,588, 609, 630, 651, 672, 693, 714, 735, 756, 777, 798, 819, 840, 861, 882, 903, 924, 945, 966, 987, 1008, 1029   } $

so your answer is $\boxed{\color{royalblue}{567}}$

By similar triangles, AD = 9/2.

first of all, welcome to webcalc 2.0!

1. solve this system of equations: x+y = 10, xy = 16, you get \(x=8, \: y=2 \). 8 - 2 = 6. keep in mind that you could do 2 - 8 and get -6 as well.

Hey there, CCeed!

Ahaha, yerpp, you're using this website correctly; thank you for creating an account so that I have someone to address as besides "Guest". 😂

1. Let's see what all the factors of 16 are: 1, 2, 4, 8, and 16

1 & 16 are a pair, 2 & 8 are a pair, and 4 & 4 are a pair.

1+16=17

2+8=10

4+4=8

x=8 and y=2 or the other way around.

Therefore the answer is 6 or -6.

As for the other two questions, I don't know how to solve them correctly...

Hope this helped! :)

( ﾟдﾟ)つ Bye

Alan pointed out to me an error in this answer ....corrected below:

log₅  24 / log₅ 2     -    log₅ 3/ log₅ 2  

  = 3     =     log₅ x

     5^3 = x = 125                (as UNTS subsequently showed below)                                          Thanx , Alan!  

oh sorry i didnt see that the first case could have zeros *smh*

73 xx       x can be 0-9        100 choices

x73y        x can be 1- 9   y can be 0 - 9    90 choices

xy 73       x can be 1-9    y can be 0-9      90 choices

total choices = 280

cases: 73 is in the front, middle or end. 

 case where it is the front: 

first - second digit, 1 choice each

3rd digit: 1~9 = 9

4th : 9

case where it is the middle:

first digit has 9 choices, and the 4th also has 9 choices. 2nd and 3rd have one choice. 

case where it is the end. 

same thing, first digit 9 choices, 2nd digit 9 choices, 3 and 4, 1 choice. 

So! 9 times 9+9 times 9+9 times 9 = ill leave that to you

I just wanted to let you know that you have not attached a graph. You can either edit the question and I can answer it or you can post a new question.

I hope this helps!!

If a = 8 (mod 9), then a must be 2 (mod 3), so we can forget about that equation. 

a = 4 (mod 5)

a = 1 (mod 7)

a = 8 (mod 9)

I'm going to use the Chinese Remainder Theorem, which is a bit difficult to exaplain, but there are some good videos on youtube. 

Starting with mod 5. 

x = 0 (mod 7) and (mod 9) but 4 (mod 5). 

x = 7*9*n = 4 (mod 5)

This works when n = 3, x = 189. 

Then, mod 7. 

x = 0 (mod 5) and (mod 9) but 1 (mod 7). 

x = 5*9*n = 1 (mod 7).

This works when n = 5, x = 225. 

Then, mod 9. 

x = 0 (mod 5) and (mod 7) but 8 (mod 9). 

x = 5*7*n = 8 (mod 9).

This works when n = 1, x = 35. 

Finally, we add everything up. 

189 + 225 + 35 = 449

We can make it smaller by doing 449 - 5*7*9 = 134. 

=^._.^=

how did you get the angles

There is only one year that ends in 00 in the interval given is 2000.

When it is divided by 900. The remainder is 200. So it is a leap year is 1.

https://web2.0calc.com/questions/help-me_35605