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√x  + √[ x + 6 ]  =  6      subtract √x from both sides

√ [x + 6 ]  =  6 - √x       square both sides

x + 6  =    36  -12√x  + x       subtract x from both sides

6 = 36 - 12√x      rearrange as

12√x  = 36 - 6

12√x =  30       divide both sides by 12

√x =  30/ 12    =  5/2      square both sides again

x  =  (5/2)^2  = 25 / 4

Suppose that  all  24  earned  a "10"    and the other 6  earned  a "6"

Then....the greatest average  is

[ 24(10) + 6(6) ]  / 30  =      276 / 30   =  9.2

The  smallest angle  =  (1) / ( 1 + 2 + 3)  * 180  =  (1/6) * 180  = 30°

The next largest angle must twice the smallest  = 60°

And the greatest must be three times the smallest  = 90°

We have  a 30 - 60 - 90  right triangle

The side opposite the 90°  angle is the greatest side and is always twice as long as the shortest side...so.......24 m

9, 13, 15, 17, 19, 23, 31, 49

The sum of these numbers  = 176

And we have 8 numbers.....so    .....the original average  =  176 / 8  = 22

And the original median = (17 + 19) / 2  =  36/2 = 18

So....the new median must be 20

So....after removing two numbers....the average is 24.....this means that the sum of the new numbers must be  24 (6)  = 144

So...the sum of two numbers removed must be   176 - 144  =   32

These pairs sum to 32

(15,17)....if we take these out...the new median is (19 + 23) /2  =  21

(13,19).....if we take these out.....the new median is (17 + 23) / 2  = 20

(9, 23)

So...the numbers must have been  13  and 19.....and their product  = 247

Here's  one method of solving this  [ graphical solution ] :

https://web2.0calc.com/questions/interesting-geometry

The common ratio   is  1/10

So.....the sum  =        first term                      0.79            0.79            7.9         79 

                               ______________   =   _______  =  _______  =  ___  =   ____

                               1 - common ratio          1 - 1/10           9/10           9           90

BTW  -  Happy Easter, WP!!!!

From the other question you posted.....we found the base  = 4

So.....the perimeter  =  3(2)  +  4    =    6  +  4   =  10  units

Draw a perendicular to the base from the top right vertex......this will form a right triangle with sides  1  - √3   -  2

And the  height = √3

And the distance from the bottom right vertex to the intersection of the perpendicular and the base  = 1

So....by symmetery.....the base length  =  1  +  2    +  1   =    4

So.....the area  = (1.2) height * ( sum of base lengths)   =

(1/2) * √3  * ( 4 + 2)  =

(1/2) (6) (√3) =

3√3  units^2

thanks, everyone!

thanks

Nevermind, the answer was that the triangle was "invalid"

You know that \(\sqrt{16}=4\) and \(\sqrt{100}=10\) so you need to find how many numbers in 17,18,19....99

My way of doing this is just making the first number 1 and the last number is the number of numbers in the set. So you need to subtract 16 from all the numbers to make the first one 1. The set becomes 1,2,3....83. Therefore there is 83 positive integers that satisfy the inequality.

We know that the median right now is 18 because (17+19)/2=18. So the median has to be 18+2 or 20. That means one of the numbers is either 17 or 19 because if you don't remove one of those numbers the median won't change. Now we take a look at are other clue. The average is increased by two. The average without removing the 2 numbers is (9+13+15+17+19+23+31+49)/8=22. That means the average after you take out the 2 numbers is 22+2=24. And that means the sum of all the numbers after you get rid of the 2 is 24*6=144 because if the average is 24, you multiply the average by the number of numbers in the set to find the total sum. Before, you get rid of the two numbers the sum is 9+13+15+17+19+23+31+49=176. 176-144=32 which means the two numbers you get rid of add up to 32. If you get rid of 17, you have to get rid of 15 because 32=17+15. But the median is then (19+23)/2=21 so you choose 19. If you choose 19 you need to also choose 13 because 32=19+13. and the median is (17+23)/2=20 so 19 and 13 are correct. The product is 247.

There is no original triangle, they just say triangle pqr

what we learned is that the side opposite to angle p is is side p (etc.)

You don’t show the original triangle, so I can only guess that you’ve got the lengths p and q interchanged.

What is 0.48 expressed as a fraction in simplest form? (just the 8 repeating)

Let y=0.4888888888.....

10y= 4.888888888...

    y=0.4888888888....

subtract

9y = 4.4

\(y=\frac{4.4}{9}=\frac{44}{90}=\frac{22}{45}\)

\(0.4\dot8=\frac{22}{45}\)

You need to check my working.   :)

For some reason the info keeps getting cut off, so:

angle p = 65

p = 18

 q= 22

Thanks, but could you explain how to solve this problem using law of sines?

In triangle PQR m

Solve the triangle (find all missing sides and angles)

Hi   Hypotenuisance

There has been a bit of discusion about your clever username.  We like it!    LOL

Welcome to the web2.0calc forum  

Because sin(65)*22/18 is bigger than 1, and there is no angle whose sine is bigger than 1.

The equation of the red dotted line is of the form y = mx + c.

It passes through points (3,1) and (1, -2), so the slope, m, is m = (1-(-2))/(3-1) or m = 3/2.

Since it passes through (3,1) we have 1 = (3/2)*3 + c, so c = -7/2

Hence The inequality describing the yellow region is y >= (3/2)x - 7/2

.

o+3.1+0.5+0.25=3.58

Make use of the identity: \(z\cdot z^{*} = |z|^2\).

\((z-5-i)(z-5-i)^* = 25\)

Let z = x + y i.

\(((x-5)+(y-1)i)((x-5)+(1-y)i) = 25\\ (x-5)^2 + (y-1)^2 = 25\)

So the equation |z - 5 - i| = 5 actually represents a circle centered at z = 5 + i and with radius 5.

Expressing the given expression with x and y:

\(|z-1+2i|^2 + |z-9-4i|^2\\ =(x-1)^2+(y+2)^2 + (x-9)^2 + (y-4)^2\\ = x^2-2x+1+y^2+4y+4+x^2-18x+81+y^2-8y+16\\ =2x^2+2y^2-20x-4y+102\\ =2(x^2+y^2-10x-2y+102)\\ =2((x-5)^2+(y-1)^2+76)\)

And because \((x-5)^2+(y-1)^2 = 25\),

\(|z-1+2i|^2 + |z-9-4i|^2\\ =2(25+76)\\ =202\)

So the minimum value is 202.

\(z^5 = 1 \implies z = e^{\frac{2k\pi i}{5}}, k = 1,2,3,4\\ \dfrac{1}{z} = z^4\\ \dfrac{1}{z^2} = z^3\\ \text{Substitute }z = e^{\frac{2\pi i}5}\\ z+\dfrac1{z}+z^2+\dfrac1{z^2} = z+z^2+z^3+z^4 = \displaystyle \sum_{k=1}^4 e^{\frac{2k\pi i}{5}}\)

So the question is actually asking the sum of roots of the equation \(z^5-1 = 0\), excluding 1.

Sum of roots of \(ax^5+bx^4+cx^3+dx^2+ex+f=0\) is \(\dfrac{-b}{a}\). But don't forget to exclude 1.

The required value = \(-\dfrac{0}{1} - 1\) = -1.

You can learn from other's solution. Come on!

https://web2.0calc.com/questions/please-help-asap_88