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Note that  triangles ADC and CDB are similar because  angle DAC = angle DCB and angle CDA = angle CDB

Thus....this implies that 

AD /DC = CD / DB

But AD = y     CD = h     and DB = x

Therefore

y / h =  h / x       and cross-multiplying we get that

xy = h^2

sorry its wrong

What kind of a triangle is it? Right triangle, acute, obtuse.....etc.?

The smallest n =14, because:

126 = 2 * 3^2 * 7 =3^2 x 14. Then if you multiply by another 14, then:3^2 x 14^2 will give you a square.

Sqrt(3^2 x 14^2) =3 x 14 =42

Post 1 question at a time and upload any picture that comes with it.

Thank you for helping me

Solve for x:
3 x^2 = 19683

Divide both sides by 3:
x^2 = 6561

Take the square root of both sides:

 x = 81   or    x = -81

The side lengths of a pentagon and two right interior angles are shown in the diagram. Find the area of the pentagon.

Assuming a and b are orthogonal then:

a.b is 0

a.c is positive

a.d is negative

b.c is positive 

b.d is positive

c.d is negative

The following diagram should help:

Note that the equation of the diagonal line is \(y=\frac{10}{17}x\)  As x goes in integer steps from 0 to 17, y only takes integer values at x=0 and x=17.

Thank you, but the answer is one of the following: 25,26,27,28

Use Pythagoras's Theorem:

Bug's Path^2 =10^2 + 17^2

                      =100 + 289

                      =389

                      =sqrt(389)

                      =19.72 - or ~20 tiles that the Bug crosses.

Three positive pairs will satisfy the equation as follows:

 a = 9, b = 153
a = 153, b = 9
a = 17, b = 17

a slightly different way of looking at it

\(\text{First choose two slots for the E's }\dbinom{5}{2}\\ \text{Then we have 3 distinct letters that we can make }3! \text{ arrangements with}\\ \dbinom{5}{2}3! = 10\cdot 6 = 60\)

Thank you, but that answer is incorrect.

One pair, because 17 is a prime number and 1+1=2 therefore a and b MUST be 17 because anything greater would yield a result less than 2/17. In addition, anything less than 17 wouldn't satisfy the equation because 17 is prime and 17 would be unobtainable as a LCD. Therefore the only pair would be (17,17). This yields the answer of only 1 pair.

Cool!

Ooops....didn'y answer all of the questions...

#2    2 moles of  AuO3  produce 4 moles of Au

      .4797 moles Au2O3  x 2 = .9594 moles Au produced

#3  85 grams of Au = 85/196.96 mole          1/2 as many moles Au2O3 is needed

      85/196.96   x  1/2 = .4315 moles

we already found Au2O3 moles weigh 441.937 gm

     .4315 mole x 441.937 gm/mole = 190.72 gms of Au2O3 are needed

x= y^2-6y+8      -b/2a     is vertex point y    =    -   -6/2 = 3  Sub in to find the x value

x = 3^2- 6(3) +8= -1

(-1,3)

See the answer here:  https://web2.0calc.com/questions/help_53845

This is equivalent to the question what positive integers less than 100 have only 2 and 5 as prime factors.

listing these isn't too hard

2, 4, 8, 16, 32, 64

5, 10, 20, 40, 80

25, 50

so 13 of them

Your class must be taking computer programming. This is a job for a simple computer program to figure out. I wrote a short code to search up to 10,000,000 numbers and came up with these:
n= 3^2 + 4^2 = 25
n= 20^2 + 21^2 = 841
n= 119^2 + 120^2 = 28,561
n= 696^2 + 697^2 = 970,225
n= 4,059^2 + 4,060^2 = 32,959,081
n= 23,660^2 + 23,661^2 = 1,119,638,521
n= 137,903^2 + 137,904^2 = 38,034,750,625
n= 803,760^2 + 803,761^2 = 1,292,061,882,721
n= 4,684,659^2 + 4,684,660^2 = 43,892,069,261,881

Circumference of larger circle = pi d =   18 pi

smaller circle = 4 pi

 After 20 revolutions the smaller circle travels 20 * 4 pi = 80 pi

    18 pi is one complete travel around the larger circle

     80 pi -   18 pi *4    = 8pi  = the tangent point

8pi /18pi  x 360 = 160 degrees.  

Cool! 

I was taking a nap !   ZZZZZzzzzzZZZZZ    Haha .....now I'm going for a run !