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180 x 1.15 =

Start with a 1 to the left of the decimal in 1.15. Multiply that 1 x 180 =180

Then continue with 1 to the right of the decimal x 180/10                   =18 + 180 =198

Then continue with 5 to the right of the decimal x 180/100   =900/100=9 +198 =207

Thanks, Knockout...!!!!

Look at the figure below

 ∠ EBA  =  ∠ DBC  =   90°

 ∠ EAB  = ∠ CDB 

So....by  AA congruency  

Δ BAE   ~  ΔBDC  

So

AB / BE  =  BD / BC  

AB * BC   =  BD * BE

EB * BD    =  AB * BC

how b****y painful.

thank you!

eh b****y h**l I missed it.

It goes in the denominator so the denominator becomes 18 rather than 9.

What is the GCF and LCM of 34 and 784? 

\(\begin{array}{|rcll|} \hline 34 &=& 2^1\cdot 17^1 \cdot (7^0) \\ 784 &=& 2^4\cdot 7^2 \cdot (17^0) \\\\ && \begin{array}{|rcll|} \hline \text{gcf}(34,784) &=& 2^1\cdot 17^0 \cdot 7^0 \quad & | \quad \text{min exponent} \\ &=& 2\cdot 1 \cdot 1 \\ &=& \mathbf{2} \\\\ \text{lcm}(34,784) &=& 2^4\cdot 17^1 \cdot 7^2 \quad & | \quad \text{max exponent} \\ &=& 16\cdot 17 \cdot 49 \\ &=& \mathbf{13328} \\ \hline \end{array}\\ \hline \end{array}\)

Here's my  best attempt :

0   =    w         ( L^2x^2     -   x^4      -  7L^4  )

         ____        ______        ___         _____

         EI6L           2                 4             60

Multiply both sides by   EI6L / w     and the lead term "disappears"

0  =   L^2 x^2     -   x^4     -  7L^4

        ______         ____      _____

             2                4            60

Multiply through by  - 60

15x^4  -  30L^2x^2  +  7L^4   =   0

15x^4  -  30L^2x^2     =  -  7L^4         factor  out 15

15 ( x^4 - 2L^2x^2)  = - 7L^4            divide both sides  by 15

x^4  - 2L^2x^2   =     - 7L^4 / 15

Take  1/2  of 2L^2  = L^2

Square this  = L^4

Add to both sides

x^4 - 2L^2 x^2  + L^4  =   -7L^4 / 15   + L^4   

Factor the left side   and simplify

( x^2  - L^2)^2  =   8L^4 / 15

Take both roots  and simplify

x^2  - L^2  =   ±√ [ (8L^4 / 15) ]

x^2 - L^2  =  ±2L^2√ [ 2/15 ]

Add L^2  to both sides

x^2  =  L^2  ± 2L^2√ [ 2/15 ]

Take both roots again

x =  ±√ ( L^2 ± 2L^2√ [ 2/15 ]  )  =    ± L√ ( 1 ± 2√ [ 2/15 ]  ) 

The Greatest Common Factor

The greatest common factor, or GCF, is the greatest factor that divides two numbers. To find the GCF of two numbers:

List the prime factors of each number.

Multiply those factors both numbers have in common. If there are no common prime factors, the GCF is 1.

The Least Common Multiple

A common multiple is a number that is a multiple of two or more numbers. The common multiples of 3 and 4 are 0, 12, 24, ....

The least common multiple (LCM) of two numbers is the smallest number (not zero) that is a multiple of both.

I typed GCF(34,784) into my calculator, and got \(\boxed 2\) as my GCF.

I did the same for the LCM and got  \(\boxed{13328}\).

http://www.mathematicsmagazine.com/applications/GCF_LCM.htm

You can use this to find both of them if you do not have a calculator that can calculate GCF and LCM.

thank you

And... if you just want a written out solution...

Suppose the six cookies to be chosen are the stars, as we attempt to implement stars and bars.

We take two dividers, and place them between the cookies, such that the six cookies are split into 3 groups, where the groups are the number of chocolate chip, oatmeal, and peanut butter cookies, and each group can have 0.

First, assume that the two dividers cannot go in between the same two cookies.

By stars and bars, there are \(\binom{7}{2}\) ways to make the groups.

Finally, since the two dividers can be together, we must add those cases where the two dividers are in the same space between cookies. There are 7 spaces, and hence 7 cases.

Our final answer is\(21 + 7 = \boxed{28}\) assortments.

Precise solution right here

https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_21

This question is from an AMC 10A math competition. I like those...they're fun.

So the question is:

Solve the quartic equation- \(2x^4 + 14x^3 + 6x^2 + 14x + 2 = 0\)

The figure below can be helpful in understanding the solution.

https://i.stack.imgur.com/2SopB.png

We are given \(\angle ACD = \angle EAC= 90^{\circ}\), so \(AE\) is parallel to \(CD\), so \(\angle EAB = \angle BDC\).

We also have \(\angle EBA= \angle CBD = 90^{\circ}\), so that \(\angle AEB = \angle BCD\).

So angles of triangles \(ABE\) and \(BCE\) are the same, hence they are similar.

Consider similar triangles \(ABE\) and \(BCD\), then we have \(\angle BDC=\angle EAB\).

For triangle \(ABE\), 

\(\tan(\angle EAB)=\frac{EB}{AB}\).

For triangle \(BCD\),

\(\tan(\angle BDC)=\frac{BC}{BD}\)

Thus we have

\(\large \frac{EB}{AB}=\frac{BC}{BD}\Rightarrow EB\cdot BD=AB\cdot BC\).

I'm not sure. I've done the same exact problem somewhere else before, but I do not recall the answer.

Here is a similar question: https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_8_Problems/Problem_20

Maybe like the question above, sort them by base lengths, and see how many triangles can be formed by connecting points from each distinct base length.

Since only the number of stickers on the sheets matters, we can list the possibilities systematically:

\(\begin{align*} & 8-0-0-0 \\ & 7-1-0-0 \\ & 6-2-0-0 \\ & 6-1-1-0 \\ & 5-3-0-0 \\ & 5-2-1-0 \\ & 5-1-1-1 \\ & 4-4-0-0 \\ & 4-3-1-0 \\ & 4-2-2-0 \\ & 4-2-1-1 \\ & 3-3-2-0 \\ & 3-3-1-1 \\ & 3-2-2-1 \\ & 2-2-2-2 \end{align*}\)

There are  \(\boxed{15}\)  possible arrangements of stickers on sheets of paper.

You want to distribute your 5 distinguishable balls into 3 indistinguishable boxes. Let \(B(5,3)\)denote the number of ways in which this can be done into exactly 3 indistinguishable non-empty boxes, and use the recurrence relation \(B(n,k)=B(n−1,k−1)+kB(n−1,k)\)with \(B(n,1)=1\) and \(B(n,n)=1\). You seek \(B(5,1)+B(5,2)+B(5,3)\).

This has something to do with "Stirling numbers" although I am not too sure on this topic. You might consider searching it up!

The Roman Empire lasted 27BC + 476 AD =503 years.

The Persian Empire lasted 5/11 x 503 =229 years

550 - 229 =321 BC - when the Persian Empire ended.

I don't think the LHS of your equation will ever equal the RHS as it is written. Can you see why?

Alright, i was hoping for something a little more detailed, but thanks for trying to help!

Yesterday morning at 8 a.m. the temperature was -14˚F. At noon it was 10 degrees warmer

So at noon it was -14˚F + 10˚F = -4˚F 

The temperature increase between noon and and 4 p.m. was twice the temperature increase between 8 a.m. and noon

So the temperature increase between noon and and 4 p.m is 10 so \(2 \times 10=20\)

Finally -4˚F + 20˚F = 16˚F   The temperature at 4 p.m!

Hope it helps! 

-15/-5=3

Now, (-5)(3)= -15

So, 3=n

:)

Hope that helped!

-SomeRandomGuyInTheInternet