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Problem 1: Since AR = AB, triangle ARB is isosceles.  Also, arc AB is 130 degrees, so angle ARB is 130/2 = 65 degrees, which means angle RAB is also 65 degrees.  Then arc RB is 130 degrees.  Since arc AQ is 28 degrees, angle RPB is (130 + 28)/2 = 79 degrees.

Problem 2: Since arc POQ is 80 degrees, angle POQ is 160 degrees.  Then major arc PQ is 360 - 160 = 200 degrees, so arc POQ on the smaller circle is (360 - 200) + 80 = 240 degrees.

Problem 3: Angle ABX = Angle AOX = 35 degrees, so Angle ABC = 180 - 35 = 145 degrees.

Would the answer be 2159.43

The above should read:

1 /7 =0.1428571428 5714285714 2857142857 1428571428 5714285714 2857142857 1428571428 5714285714 2857142857 14285714285714 = 8 is 100th digit.

Sorry abou that.

1 l s  / 5 l f   x  2 l f  = 2/5 l s

Expand L side

5 + g/8 = 6g+8 - 6g

5 + g/8 = 8    subtract 5 from both sides

g/8 = 3         Now, Multiply both sides by 8.......

1 /7 =0.1428571428 5714285714 2857142857 1428571428 5714285714 2857142857 1428571428 5714285714 2857142857 1428571429

1. Angle RPB is 72 degrees

2. Arc POQ is 220 degrees.

3. Angle ABC is 135.5 degrees.

yes it is 

954 - 459 = 495

The area of AEFGD is 64.

The radius of the cone is 14/sqrt(3), and the height of the cone is 28, so the volume of the cone is 1/3*pi*(14/sqrt(3))^2*28 = 5488*pi/9.  To the nearest whole number, this is 1916.

Let the circumference of the rear wheel =C
The circumference of the front wheel     =C - 6
360 / [10*(C - 6)] =360 / C, solve for C
C = 6 2/3 - meters - circum. of rear wheel.
6 2/3 - 6 =2/3 - meters - circum. of front wheel.

I think teaching answers are usually much better.....    So long as the answerer actually knows what they are talking about. 

If the asker have already attempted it they should show their own working,  and they should ask if someone can find their mistake.

For some people whole answers are good but there are far too many people here who are interested only in the answer and working that they can copy. By giving them the whole answer they will only learn a method to cheat.  They will learn no maths.

Even for good kids, they will usually learn more if they have to do some of the thinking for themselves.

Remember   log (a/b)  = log a - log b      Does that help?

At quick glance I think Q2  is T  F  T 

I think thisis the pic?

Example Albert:

$1000 earned 1.2% annual interest compounded monthly          =   1000 (1+.001)120  

                                                                                                                 (periodic interest = .012/12 ,n is periods = 10yr x 12 mos)

$500 lost 2% over the course of the 10 years                                =    500 (.98)

$500 grew compounded continuously at rate of 0.8% annually     =  500 e^.008(10)        10 years    interest .008 (in decimal form)

   Add these three to see how Albert did with his investments

Then A' would be on BC which the question states is not allowed.

                  3x^2  - 7x  -  6          (3x + 2) ( x - 3)             3x +  2

f(x ) =        ____________ =     _____________  =       _______       (1)

                  2x^2  - 5x  -  3          (2x + 1) ( x -3)              2x  +  1

The x coordinate of the   "hole"  can be  found as    x  - 3  = 0  ⇒   x  = 3

To find the y coordinate of the hole......sub  this x value into  (1)   and we have

3(3)  + 2              11

_______  =        ____ 

2(3) + 1                 7

So....the "hole" occurs at    (3, 11/7)

See the graph, here :

https://www.desmos.com/calculator/7mdt3lommd