All Answers 358187 Answers

It is a mathematical EXPRESSION (NOT an equation), and as such , it can be ANY value.

IF you meant to ask : "What is  5x + 2 =0 ? " , then you have an EQUATION that can be solved thusly:

5x+2=0        Subtract 2 from both sides of the EQUATION

5x = -2         Divide both sides by 5

x = -2/5

Similarly, in triangle ACD, angle DAC=33.2 deg, angle DCA=66.4 deg, and angle CDA=80.4 deg.

Side AC=4.  By Law of Sines: CD=2.221 and AD=3.718. Therefore:

AD^2 =3.718^2 =13.823 ~ 1,120/81.

The probability that none get it right is q = (1-0.25)*(1-0.5)*(1-0.6)*(1-0.8) or q = (3/4)*(1/2)*(4/10)*(2/10) → 3/100

Hence the probability that at least one gets it right is p = 1 - q or p = 1 - 3/100 → 97/100

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log₃ 2187

We can use  the change of base theorem, here

The answer is given by

log 2187  / log 3  =    [  the onsite calculator gives  6.99999.....]  

The actual answer is 7

To see this....we are asking....What exponent do we put on 3 to get 2187 ??

So

3⁷  = 2187

Answered here :

http://web2.0calc.com/questions/help_90006#r1

I think I answered the second one here :

http://web2.0calc.com/questions/help_49408#r1

First one

12x^2  - 64  =  -x^4     add  x^4  o both sides

x^4  +  12x^2  -  64  =  0     factor as

(x^2  + 16)  (x^2  - 4)  = 0

(x^2  + 16)  (x + 2) (x - 2)  =  0

We only have 4 solutions.....  setting the last two factors to 0 and solving for x  we get that x  = 2   and x  = -2

So....the first answer is correct

7%  increase  per year  =  (1.07)...so.....

f(x)  =  295  (1.07)^x

After 2 years

f (x)  =  295 (1/07)²    =  338

y  =  487 (1/2)^{(1/36 * x)}

After  x  =   5 days

y  =  487  (1/2^{)(1/36 * 5})  ≈ 442.302  kg

(-3i) (-4i)  =

-3 * - 4  * i * i   =

12  * i^2           ⇒ [ i^2  = -1  ]

12 * -1  =

-12

x + y  =  5   ⇒   y  =  5  - x     (1)

x^2  + y^2  =  25     (2)

Sub  (1)  into (2)

x^2  +  ( 5 - x) ^2   =  25

x^2  + x^2  - 10x  +  25  = 25   subtract 25 from both sides

2x^2  - 10x  = 0    factor

2x ( x  - 5)      set both facoters to 0 and solve for x and x  =  0   or  x  = 5

Using (1)

When x  = 0,     y  =  5 - 0   =  5

and

When  x  = 5,  y  = 5 - 5  =   0

So the solutions are    (0, 5)  and  (5, 0 )

First one, NSS

5^3x  =  900  take the log of both sides

log 5^3x  =  log 900      and we can write

(3x)  log 5  =  log 900     

x  (3log 5)  = log 900     divide both sides by 3log 5

log (900) /  [ 3log (5) ]  ≈ 1.41

ln 3 +  ln (4x)  =  4    we can write

ln (3*4x)  =  4

ln (12x)  =  4

This says that

e^4  =  12x         divide both sides by 12

e^4 / 12  = x ≈  4.55

Thanks so much!!

See the first graph here, NSS :  https://www.desmos.com/calculator/jsrz1zskfu

[ The feasible region  is bounded by the three points shown ]

Here's the graph of the second one :  https://www.desmos.com/calculator/kfhsp57phm

y  =  l 8x  - 2 l  - 7    

We can solve this....if the solution is positive, the the shift is to the right....if the solution is negative, the shift is to the left

8x  - 2  =  0     add 2 to both sides

8x  = 2            divide both sides by 8

x =  2/8  =  1/4  =   1/4 units to the right  =  axis of symmetry

vertex (1/4, -7),   axis of symmetry  = x  = 1/4     1/4 to the right  and  7   down

point  (11, 3)   slope  4/11

y  - 3   =  (4/11) (x  -  11 )

-10x  +   11y   = -  6    add 10x to  both sides

11y   =   10x  - 6       divide both sides by 11

y  =  (10/11)x  -  6/11

slope  =  (10/11)     y int  = - 6/11

4x  +  15  ≥  - 9    subtract 15 from both sides

4x ≥ - 24         divide both sides by 4

x ≥  -6

8x -  6   ≤  34      add 6  to both sides

8x ≤ 40    divide both sides by 8

x ≤  5

l 4x  +  3  l   =   3           we have two equations

4x  +   3   =    3                                     4x  +  3   =   - 3

                                      subtract 3 from both sides

4x   =  0                                                4x =  - 6

                                       divide both sides by 4

  x  = 0                                          x  =  -6/4  =  -3/2

y = ax^2 + bx  +  c

The x coordinate of the vertex  =  -b / [2a]  =  2  ⇒  b = - 4a    

We know that

a(2)^2 + b(2)  +  c  =  3  ⇒  4a  + 2b  +  c  =  3  ⇒  4a -  8a  + c  =  3  ⇒  -4a + c = 3    (1)

a(4)^2 + b(4)  +  c  =  4  ⇒ 16a  + 4b  + c  = 4  ⇒  16a - 16a + c  = 4  ⇒  c  = 4   (2)

Subbing (2)  into (1)  we have that   

-4a  +  4 =  3

-4a  =  -1

a  = -1/-4  =  1/4

And     b  =  -4(1/4)  =  -1

So......the function is 

y  = (1/4)x^2  -  x  +  4

So....when  x  = 6 

y  =  (1/4)6^2 - (6)  +  4  =  9  - 6  +  4   =   7

In triangle ABD, angle BAD =33.2 deg., Angle ABD=47.2 deg. Angle ADB =99.6 deg.
Side AB = 5
Using the Law of Sines, we get:BD=2.777 and AD=3.721, so:
AD^2 =3.721^2 =13.846 ???!!
CPhill: Does this make sense???

\($\frac{x}{y} = \frac{225}{xy} + \frac{y}{x}$\)

Rearrange as

x / y  -  y / x  =   225 / xy

[ x^2  -  y^2 ]  / xy  =   225  / xy

x^2   -  y^2   =   225

( x + y) (x - y)   =  225

The factors of 225  are :  1 | 3 | 5 | 9 | 15 | 25 | 45 | 75 | 225 (9 divisors)

So......the possible pair combos are

(x +y) (x - y)  =   225

225       1

75         3

45         5

25         9

15        15    

This appears to set up the following  systems  of equations :

x +  y   = 225

x -  y   =  1        add  these  and    2x  =  226   → x  =  113   and y  =  112

x +  y   =  75

x +  y  =    3    add these and  2x  =  78  →  x  =  39    and y  =  36

x +  y  =  45

x  -  y   =   5   add these and   2x  = 50  →  x  = 25   and y  =  20

x + y  =  25

x - y  =    9    add these  and  2x  =  34   →  x  = 17   and y  =  8

x +  y   =  15

x -  y  =    15    add these  and  2x  = 30  →   x=  15  and y  = 0

So  for  x, y > 0......the  pairs  are

(113, 112)   (39, 36)  ( 25, 20)   and  (17, 8 )

Probably more elegant ways to do this.....but....

Here's a diagram :

AD  will bisect BAC.......and this angle bisector  creates the following ratio

BA / AC  =  BD / DC

5/4  =  BD / DC

Let  DC  =  x     then BD  = (5/4)x

And  DC + BD  = 5        so

x + (5/4)x  = 5

(9/4)x  =  5

x  = 20/9  =  DC

Draw  altitude  BG to AC .....this will bisect AC....   and  draw DF perpendicular to AC

Now......BG  =  √[BC^2 - GC^2]  = √[ 5^2 - 2^2]  = √21

And   triangles BGC  and DFC  are similar

And  BG/ BC   is similar to  DF/DC......so

√21 / 5  =  DF/ (20/9)     so

(√21 / 5) * (20/9)  =  DF   = 4√21 / 9  

Likewise

GC /  BC  =  FC /DC

2/5  =  FC / (20/9)

(20/9)(2/5)  = FC  =  40/45  = 8/9

And AF  =  AC  - FC  =   4  -  8/9     =  28/9

But  ADF  forms a right triangle  with   AF^2  + DF^2   =  AD^2

AF^2  =  (28/9)^2  =   28^2 / 81  =  784/81

DF^2  = [ 4√21 / 9 ]^2  =   16*21 / 81  =  336/81

So    AD^2  =    [ 336 + 784 ] 81  =  1120/81

simplify | 6/(sqrt(36) + sqrt(27)) + 6/(sqrt(27) + sqrt(18)) + 6/(sqrt(18) + sqrt(9))

6/[6 + 3sqrt(3)] + 6/[3sqrt(3) + 3sqrt(2)] + 6/[3sqrt(2) + 3]

2/[sqrt(3) + 2] + 2/[sqrt(3) + sqrt(2)] + 2/[sqrt(2) + 1]

4 - 2 sqrt(3) - 2 sqrt(2) + 2 sqrt(3) + 2 sqrt(2) - 2

= 2

Thanks! but that is incorrect, you forgot 1

a=1 and x=1

a=5 and x=5

a=7 and x=7

a=11 and x=11

These are solutions for "a" < 12.

You are right. Sorry about that. [Edited].

If the axis of symmetry is x  = -3, then  - 3 is the x coordinate of the vertex

And....the x coordinate of the vertex is given by    -b / (2a)

So we have that

-b / (2a)  =  -3

b / (2a)  = 3       multiply both sides by 2

b / a   =   6

Thanks!