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Another method:    x = 585/z       y = 630/z

xy = (585 * 630) / z² = 182       so z = 45        

now you can find x   and y      xz = x(45) = 585    soooo x = 13       etc

Here are the last 10 digits of:

123^123 mod 10^10 =....0220839267  and as you can see 7 is the units digit.

Solve xz = 585, xy = 182, yz = 630.

1. Multiply them all together    (xyz)² = 67076100

2. Take square root  xyz = 8190

3. Divide xyz by yz:   x = 8190/630   x = 13

4. Divide xyz by xz:  y = 8190/585    y = 14

5. Divide xyz by xy  z = 8190/182     z = 45

15 -7  = 8% difference = $10

8% x = 10

.08x = 10

x = $125

Jenny wrote 5x instead of 4x

You didn't include the parent funtion in your question, so I cannot sketch it   BUT

   I can tell you   #1   the graph will be shifted one unit to the LEFT

                           #2  the graph will be shifted   5 units DOWN

Jeremy bought 2 gumball refills @ $4.95 each. So:

2 x $4.95 =$9.90 - cost of 2 refills.

$19.95 for gumball machine + $9.90 for 2 refills =$29.85 -  amount spent.

Because Jeremy spent MORE than $25 on merchandise (I assumed that 2 gumball refills are merchandise !!), then he must pay $6.95 in shipping and handling.

$9.90 + $19.95 + $6.95 =$36.80 - what Jeremy paid in total.

Here is a pic to help you.

If 1-cos(angle) / sin(angle) = sqrt3 / 3,
what is angle? show proof

Hello Guest!

\(1-\frac{cos (x)}{sin (x)}=\frac{\sqrt{3}}{3}\)           \(1-\frac{cos (67.732deg)}{sin (67.732deg)}=\frac{\sqrt{3}}{3}\)

                                       \(1-0.42265=0.57735\)

\(\frac{1}{1-\frac{cos (x)}{sin (x)}}=\frac{3}{\sqrt{3}}\\ \frac{1}{1-cot(x) }=\sqrt{3}\\ 1=\sqrt{3}-cot(x)\cdot\sqrt{3}\\ cot(x)\cdot\sqrt{3}= \sqrt{3}-1\)

\(cot(x)=1-\frac{\sqrt{3}}{3}\)

\(angle\ (x) = 67.732\ deg\)

  !

Find the angle x.

ABCD is a parallelogram. E is a point on AB such that AE:EB = 1:2, F is the mid-point of CD. AF meets DE at G and BF meets CE at H. When expanded on both sides, GH meets AD at M and BC at N. If the area of ABCD is 420, find the sum of the areas of the triangles ∆CNH and ∆DGM.

The sum of the areas of the triangles ∆CNH and ∆DGM is 48 u²

n⁸/n⁵ = 512

n^(8-5) = 512

n³ = 512

n = 8

How can it be 735 if the area of a whole parallelogram is 420 units squared?!?

We shall start by prime factorizing 2015. It is equal to 5*13*31. We want to minimize our number, so we must minimize it's prime factors. This means that our answer is (2^30)(3^12)(5^4)

Lowest is 3+4+5=12. Highest is 101+102+103=306. You can obviously get every number in between by increasing some variable or decreasing another. So ans is {12,13,14,...,306}

2*9+2*4=18+8=26

If first is x-2, second is x, third is x+2. (x-2)+x+8=3(x+2)-11 --> 2x+6=3x-5 --> x=11. The third is 11+2=13

By similar triangles 2x+25=(14+x)/2 --> 4x+50=14+x --> 3x+50=14 --> 3x=-36 --> x=-12

GE = 14+(-12)=14-12=2

a--> 3y=36x+24 --> y=12x+8 --> slope=12

12=3b --> b=4

c=b/2 --> c=4/2 --> c=2

Add first two --> 3x-2z=5 --> 3x=2z+5 --> x=(2z+5)/3

Add second and third --> 3x-y+3z=3 --> 2z+5-y+3z=3 --> 5z-y=-2 --> y=5z+2

Use last eqn. --> (2z+5)/3+5z+2+2z=-4 --> 2/3z+5/3+7z=-6 --> 23/3z=23/3 --> z=1

So y=5*1+2=7, and x=(2*1+5)/3=7/3

Ans: (7/3, 7, 1)

12=2^2*3^1 --> has (2+1)(1+1)=3*2=6 factors, in positive. By symettry, this is the same for negative. So 6+6=12.

The formula is \(S_{n}=\frac{(T_1+T_n)n}{2}\). Note that t_12=(26+2)-(2*12)=28-24=4. Note that t_1=26. So we have \(S_{12}=\frac{(4+26)12}{2}=\frac{30*12}{2}=30*6=\boxed{180}\)

It is not. It doesn't pass the horizontal line test at any value for y, where y<=-5.

https://www.desmos.com/calculator/nwmsbaxflf

You have made some good progress. BDC is a 30-60-90 triangle. This means that BD:DC:BC=1:sqrt3:2. It also means that angle CBD=60 degrees. So our answer is 45+60=105 degrees