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Amen .

All three of those expressions must add to 180

x+22       +     x+6       +     x-28     = 180

3x = 180

x =  60                                  then the angles are    82      66   and 32 ^o

^o

It is 72.   There are 5 less than 72 and 5 more than 72.   A median is the number that there's the same amount of numbers on both sides of it.  . 

x + 37 +7x - 9 = 180

8x = 152

x = 19

Area = (1/2)  ( AB) ( BC)   sin 30

35 =   (1/2)  ( x + 3) ( x -1)  (1/2

35  =  (1/4)( x + 3)  ( x -1)    multiply through  by 4

140  =  (x + 3) ( x -1)

140   = x^2   +   2x  - 3        rearrange as

x^2 + 2x    -143  =  0        factor as

(x - 11)  ( x + 13)   =  0

The first factor  set to  0   gives us what we need

x - 11   =   0

x  =11

AB  = x + 3   =14

BC  = x  - 1   = 10

AB  + BC    = 24

We cannot  have  the denominator  = 0

So....to  find  the values of   x  that  make it 0,  we have

2x^2  - 6x    +   4     =  0       divide  through by 2

x^2   -3x  + 2  =  0      factor

(x - 2)  ( x - 1)  =  0

x  -2  = 0                    x -1  = 0

x  = 2                            x  =1

A  + B  =   2  +  1   =   3

There are 7*48*10 = 3360 possible four-digit numbers.

We can use coordinates.  A = (0,sqrt(200)), B = (sqrt(200),sqrt(200)), C= (sqrt(200),0), D = (0,0).  Then G = (sqrt(200)/2,17), E = (sqrt(200),34), F = (sqrt(200),17), so the area of triangle GCD is 45.

The   x/2   means   that   the  domain  of g (x)  is  twice as wide as that of h(x)  [ it  horizontally "stretches" the domain of  h(x)  by a factor of 2]

So

[ -16, 16 ]    = domain of  g (x)

3sqrt (2)   =  sqrt (18)

Using  the  secant tangent theorem

AP^2   = AB  ( AB + BC)

Since  AB  =BC   then  let them  =  x

AP^2  =  x  ( x + x)

(sqrt 18)^2   =  x (2x)

18  = 2x^2     divide both sides  by 2

9 = x^2

3  = x

So

AC  = 2x   = 2 (3)    =  6

Look  like   4 + 3  + 2  + 1   =    ????

You should ask only one question at a time.

sqrt a  / sqrt b  -    sqrt b  /  sqrt  a   =

(sqrt a)^2   -  (sqrt b)^2                a   -   b

___________________   =        _______       (1)

     sqrt (ab)                                 sqrt  (ab)

(a - b)^2   =   a^2  - 2ab  + b^2     =   ( a^2 +  b^2)  -  2ab    =  ( 4ab)   - 2ab  =  2ab

So

(a - b)^2   =  2ab         take the square root of  both sides

sqrt  [ (a -b)^2]   =   sqrt (2ab)

a  - b   =   sqrt (2ab)

So, subbing  into (1)

sqrt (2ab)              sqrt 2   * sqrt (ab)

_______  =           ______________   =      sqrt (2)

sqrt  (ab)                   sqrt (ab)

-10x +8y=-44

4x+8y=40

First, let's subtract these two equations

If we do that we get:

-14x=-84

divide both sides by -14 

x=6

We can now plug in x=6 in either of the two equations. I'll plug it into the second one

4(6) + 8y = 40

24+8y=40

-24        -24

8y= 16

---    ----

8       8

Y=2

a =  log 15 / log 3

b =  log 15  / log 5

So

1/ ( log 15 / log 3)   +   1 / ( log 15 . log 5)   =

log 3  / log 15    +   log 5   / log 15         =

(log 3  +  log 5)   / log 15                  using a log property

log (3 * 5 )  / log 15

log 15 /  log  15    =     

1

well, since Carlos planted 12 plants last year, 1/3rd of that is 4, and 4 + 6 is 10.

so, i think the answer this 10.

hope this helped! please correct me if i am wrong:)

ab = 1

bc = 2

cd = 3

de = 4

ea = 5

Divde  the  second equation by the first

c/a  = 2      ⇒  c = 2a  ⇒  c/2   =a

Divide the fourth equation by the third

e/c = 4/3   ⇒  e  = (4/3)c 

Sub  these into the fifth equation

ea  = 5

(4/3)c * c/2  =   5

(4/6)c^2  =  5

(2/3) c^2  =  5

c^2  =  (5 * 3)  /2  =  7.5

c =  sqrt (7.5)

e  =  (4/3) sqrt (7.5)

a = sqrt (7.5) / 2

b =  2 /(sqrt (7.5)

d  = 3/ sqrt (7.5)

I think its 6. I need answer now!!!!

I believe that the first equation leaves you with -7 = -7 for any number "n". (This has infinite solutions)

The second equation leaves: -15 = 10. No value of "n" works for this. (This has no solutions)

The third equation leaves: -3 = 10. (This also has no solutions)

The last equation leaves you with n = 2. (This is only one solution.)

That's very crafty, guest......excellent   !!!

I definitely would not  have seen that  method   !!!!

An equation can have infinite solutions if the variables go away and you are left with something like "5 = 5" or "7 = 7".

Here's  the  key

Simplify  both sides

If  one side  = the other side, we have infinite solutions

(Hint : look at the  first one)

The sum of the numbers from 1 to n is \(\frac{n(n+1)}{2}\).

In this problem, \(n = 10^{2016}\).

Instead of multiplying everything out, we could realize that \(10^{2016}\) is even, so \(10^{2016}+1\) is odd (and therefore will not have any 2's in its prime factorization).

Now we only need to find the number of 2's in \(10^{2016}\), and get rid of one of them because we divide by 2 in the fraction \(\frac{(10^{2016})(10^{2016}+1)}{2}\).
\(10^{2016} = 5^{2016} * 2^{2016}\), so we now know that there are 2016 twos in \(10^{2016}\). We divide the expression by 2, so one less than 2016 is which 2015.
So, there are 2015 twos in the prime factorization of the sum, and it is our answer. I hope this helps.