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I won't directly solve this problem, but hint: try using sin on triangle RPM. You don't need theta, just pull up a chart of all sin values and compare the results. After that, use theta to calculate side d.

R==Roses,  S==Sunflowers,  T==Tulips

R/S ==5/6

S/T ==6/9

3/4 =[R - 50] / S, solve for R, S, T

Use substitution to get:

R ==500 Roses before the sale of 50 Roses

S ==600 Sunflowers

T ==900 Tulips

500 Roses  -  50 Roses ==450 Roses left after the sale.

Check - It's incorrect.

I think you kind of have an idea of what this problem is about based on the title of your question. Try using the law of cosines first.

nvm found the answer

will anyone help? please?

I need help putting this answer into an interval.

Since it's already 6 hours, might as well hint a bit.

My latex is a bit sluggish so forgive the frac sitting there.

\(x=\frac{-(-2)\pm\sqrt{4-4/frac\sqrt{3}/2}}{2}\)

here, somebody has answered this question before: https://web2.0calc.com/questions/help-math_22

hope this helps!

already answered

https://web2.0calc.com/questions/geometry_22332

so we can conform the "after" ratio to fit the amount in the first ratio.

Since we know that T(tulips)₁=T_2, and S(sunflowers)₁ = S₂, we find that the ratio of S and T from the first ratio, determined by 9/6 and 6/4, is 1.5. We take the ratio of roses and multiply it by 1.5,and we get 4.5. therefore, 50 roses is 10% of the total. we divide 4.5 by 0.5, getting 9, and multiply that by 50.

9*50=450

the answer is 450 roses.

thank you for the hint! i originally tried solving the problem like this, but i got stuck, haha. i'm probably just not good enough 

I would either try to get (x²-4x+2) to 1, which would maintain any power applied to it to have an output of one, or getting the exponent (x²-5x+2) to 0, so that it would constantly be an output of one. Of course, I don't know exactly how to solve this problem, but hopefully this helps. 

The domain of g(x) is [-2,1].

The probability is 1/4.

The probability is 1004/2009.

What is the sum of all possible values of for which x^2 + kx - 10x + 36 is the square of a binomial?   

For the square of the binomial to end with +36, the binomial must be either (x + 6)² or (x – 6)². 

So (kx – 10x) has to equal either +12x or –12x. 

If (kx – 10x) equals +12x, then k is +22. 

If (kx – 10x) equals –12x, then k is –2. 

The sum of +22 and –2 is 20.  

_.

In triangle PQR, PQ=13, QR=14, and PR=15.
Let M be the midpoint of QR.
Find PM.

cos Rule:
\(\begin{array}{|rcll|} \hline 15^2&=&13^2+14^2-2*13*14*\cos(Q) \\ 2*13*14*\cos(Q) &=& 13^2+14^2-15^2 \\ 2*13*14*\cos(Q) &=& 140 \quad | \quad : 2 \\ 13*14*\cos(Q) &=& 70 \qquad (1) \\ \hline \end{array}\)

cos Rule:
\(\begin{array}{|rcll|} \hline PM^2 &=& 13^2+\left(\dfrac{14}{2}\right)^2-2*13*\dfrac{14}{2}*\cos(Q) \\\\ PM^2 &=& 13^2+\left(\dfrac{14}{2}\right)^2-13*14*\cos(Q) \\\\ 13*14*\cos(Q) &=& 13^2+\left(\dfrac{14}{2}\right)^2-PM^2 \\\\ 13*14*\cos(Q) &=& 13^2+7^2-PM^2 \\\\ 13*14*\cos(Q) &=& 218-PM^2\qquad (2) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1)=(2):& 70 &=& 218-PM^2 \\ & PM^2&=& 218 - 70 \\ & PM^2&=& 148 \\ & PM^2&=& 4*37 \\ & \mathbf{PM}^2 &=&\mathbf{ 2\sqrt{37} } \\ \hline \end{array}\)

t is not in the domain when the denominator equals 0. 

When does (t-1)^2 + (t+1)^2 = 0?

=^._.^=

If \(x = \pm \infty\) then \(g(x) = 0\) so a = 0

If \(x = 0\)  then \(g(x)=1 \)  so b = 1

First work out what the roots of the first one are.

Use the quadratic formula.

Come back with your answer.

I have not looked at your question or answer, 

BUT it is great to see that you have answered for yourself and shared your answer with others    

Find the length of the zip line. (Round your answer to the nearest ft.)

Hello Guest!

You calculated correctly. The height of the tower is 300 ft.

Now follows Pythagoras:

\(z^2=200^2ft^2+300^2ft^2\\ z=\sqrt{(200^2+300^2)ft^2}\\ z=\sqrt{(130\ 000)ft^2}\\ \color{blue}z=360.555ft\)

The length of the zip line is 361ft.

  !

A Shopping mall has an area of 3.2cm^2 on the map.

calculate the are of the actual shopping mall in square kilometres.

welp, i just answered my own question

Since we need to transform 13x^2+nx-17 into (ax+b)(cx+d), we know that 13x^2+nx-17=axcx+axd+bcx+bd. 
ac=13, nx=axd+bcx=x(ad+bc) and -17=bd and n=ad+bc
Since 13 is prime, a or c has to be 13 or 1. 
The same thing goes for bd, since -17 is prime, b has to be 17, 1, -1, or -17 and if b is 17, d has to be -1, if b is -17, d has to be 1, if b is 1, d has to be -17 and so on.
plugging a and c in makes it either
(x+b)(13x+d) or (13x+b)(x+d) and plugging b and d in makes it either
(x+17)(13x-1) or (x-17)(13x+1) or (x+1)(13x-17) or (x-1)(13x+17).
For (x+17)(13x-1), we can expand it to 13x^2+220x-17, therefore n can equal 220
For (x-17)(13x+1), we can expand it to 13x^2-220x-17, therefore n can equal -220
For (x+1)(13x-17), we can expand it to 13x^2-4x-17, therefore n can equal -4
For (x-1)(13x+17), we can expand it to 13x^2+4x-17, therefore n can equal 4
From this, we can see that n=220, -220, -4 or 4.