All Answers 358501 Answers

x = mom age now    t = Tims age now

(x-5) = 3(t)

and

x + t =45

   t = 45-x     sub this in to the first equation

(x-5) = 3(45-x)

x-5 = 135 - 3x

4x = 140

x = 35= Mom's age now

Let x = .51111111....

10x = 5.11111111

100x= 51.1111111

100x - 10x = 90x   = 51.1111111...-5.11111111.....

                     90x = 46

                         x = 46/90 = 23/45

..

Already posted/answered

Put in -3   wherever there is an 'x' and calculate the result....

  Need more help than that?

Find the LCD of the fractions in the equation  9 & 3    

27 is the LCD....multiply both sides of the equation  by 27 and simplify

   Need more help?     

thank you 

Just replace 'x' with -2 and calculate...

(-2)^3 -2(-2) = f(-2)

-8 +4 = -4

3/4 x + 3 = 7/10      Find the LCD of the fractions..... 20 in this case    Multiply through by 20

(20) 3/4 x + 3(20) = (20)7/10

15 x +60 = 14

Tan 17degrees = .3057 = 9/x      x= 29.44

Multiply through by the common denominator of 3, 4 = 12

So we have

(12)(2/3)x  - (12)(2) = (12) (3/4)

8x - 24  =  9

If we draw a perpendicular from the top right vertex to the lower base......we will create a 30-60-90 right triangle.....the right side of the trapezoid will be the hypotenuse = 2

And the height of this triangle = side opposite the 60” angle = sqrt(3)  =  height of trapezoid

The angle opposite the 30 degree angle at the top  = (1/2)hypotenuse = 1

Using symmetry....we have the same situation on the left side of the trapezoid

So...the perimeter =    2 units on  the top base.....2 + 1 + 1 = 4 units on the bottom base

And 2(2) = 4 units on the sides

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

Area   =   (1/2) height of trapezoid ( sum of base lengths)  =

(1/2)sqrt(3) *  ( 2 + 4)  =

(1/2) sqrt (3) (6)  =

(3)sqrt (3)    units^2

THANKS TO EP FOR POINTING OUT MY ERROR IN THE HEIGHT CALCULATION.....!!!!

The important values for c are -4, -1, 0.  Now what would you do?

10 x (3 -- 4 )    = 30 to 40 minutes

It  IS "D"     !!!!

Good job !!!!

it's D

\(P[\text{at least 1 yellow bull in bag 1}]=\\ 1-P[\text{no yellow bulls in bag 1}] = \\ 1 - \dfrac{\sum \limits_{w=3}^{10}\dbinom{10}{w}\dbinom{8}{11-w}}{\dbinom{32}{11}}=\dfrac{206719}{206770}\)

I'll post #2 when I figure it out

\(P[\text{white bull in group of leftovers}] = \\ 1 - P[\text{all white bulls in bag 1 or bag 2}] = \\ \text{(think of bag 1 and bag 2 as one big bag here)} \\ 1 - \dfrac{\sum \limits_{y=4}^{12}~\dbinom{14}{y}\dbinom{8}{12-y}}{\dbinom{32}{22}}=\dfrac{2456369}{2481240}\)

For the first one...make sure your calculator  [ or our calculator] is set to "degree" mode!!

1.  First of all ...arcsin (-12)  doesn't exist

I think you mean   arcsin (-1/2)  = -30°   = [360 - 30 ]° =  330°  [ positive result ] 

2. arccos(3√2).....I think you mean arccos (√3/ 2)...set your calculator to "radian" mode

arccos(√3/2) ≈ .523 rads

Note.....that on the second one...if we want the exact result...set your calculator back to "degree" mode and evaluate arccos(3√2)  =   30°

To convert this to an exact result....use this

30° x    pi / 180° =   (30/180) pi  = pi/6   (rads)

Just use a calculator to find  (unless you know )

   arcsin (-1/2) = -30 or  210 degrees

arc cos (sqrt2 *3) is invalid

    Arccos(sqrt(3)/2)   is also 30 degrees which is pi/6 radians   ( or - pi/6 radians also)

Here is the graph:

Expand to x^2+7.5x - 38.5 > 0    Use quadratic formula to find x = -11  and 3.5

These are the zeros of a normally upright parabola.....so x outside of these numbers, the fxn would be > 0

     BETWEEN ( & including the values) the values of the function are zero or negative

so          -11 >x>3.5

If you use the quadratic formula to solve for this quadratic....you will see that when the term

  sqrt(b^2-4ac)   is equal to ZERO , you will have only ONE x root

so   n^2 - 4(1)(9) = 0       Solve for n

        n^2 = 36

       n = 6

Awesome! I totally forgot about that XD

thx again!

See here:

https://web2.0calc.com/questions/please-help-10_2#r2

These questions are from AoPS, a math site. I am doing it, and these are the homework problems. AoPS specifically said not to post any of these questions on the site, so just saying.