CPhill

              A

       3           5

B           D            C 

              7

Law of Cosines

7^2 = 3^2 + 5^2  - 2(3*5)cos ( BAC)

[7^2 - 3^2 - 5^2 ] / [ -2 * 3 *5] =  cos (BAC)  =  -1/2

arccos (-1/2)  = BAC =  120°

So  cos (BAD)  =   cos (120 / 2) =  cos 60 =  1/2

(a  + 2) / ( a + 1)  = b /3     cross-multiply

3(a + 2) = b (a + 1)

3a + 6  =  ab + b

3a - ab - 1b  + 6  = 0       multiply   the coefficients on a,b = (3)(-1) = -3  and add to both sides

3a - ab -1b + 6  - 3 = -3       

3a - ab - 1b + 3   =  -3

(-a -1)(b -3)  =  -3

- (a + 1) ( b -3)  = -3

(a + 1) ( b -3) =  3

a     b

0     6

2     4

-2    0

-4    2

1.   Some possibilites are

        98 , 89

        97,  79

        96,  69....etc.

2. 

D = Dad's age now     G = Grogg's age now

 D - 7  = 9( G - 7)   →   D - 7  = 9G - 63  →  9G - D  = 56     (1)

      D - 4 = 5(G - 4)  →   D - 4 = 5G -20  → 5G - D  =  16    (2)

Multiply (1) by 5   and (2)  by -9

45G - 5D = 280

-45G + 9D = -144      add these

4D  = 136

D = 136 / 4  =   34

(n + 2) / n   *  (n + 3)/ ( n + 1) * ( n + 4) / (n + 2)  = 10     simplify

1/n * (n + 3) / (n + 1) * (n + 4) =  10

(n + 3) (n + 4)  =  10 ( n)(n + 1)

n^2 + 7n + 12 = 10n^2 + 10n

9n^2 + 3n - 12 = 0 

3n^2 + n - 4  =  0

(3n + 4) (n -1 )  = 0

n = 1      n  =  -4/3

Only n = 1  is good

Value of  first fraction  =  3 / 1  =  3

The common ratio is  .036 / .4 =  9/100

The sum is

.4 / (1 - 9/100)  =  (4/10) / ( 91/100)  =  40 /  91

Slope =  [ 17 - 8 ] / [ -4 -5 ] = 9/-9  = -1

Equation of line using (5,8)

y = -1 (x -5) + 8

y = -x + 13

x + y  = 13      divide through by 13

x  / 13  + y /13  = 1

a = 13

Let  d  be the distance between  the center of the circle and UV

We can  form a right triangle  with legs of d , (UV/2)   and a hypotenuse of r

So  we have

d^2  + (UV/2)^2  = r^3

d^2  + 32^2  = r^2      (1)

Likewise  let  the distance between the center of the circle and  YZ = d + 2 + 2 = d + 4

And similar to  the first case we have

(d + 4)^2 + (YZ / 2)^2  = r^2

(d + 4)^2 + 24^2 = r^2    (2)

Set (1)  = (2)

d^2 + 32^2  = (d + 4)^2 + 24^2       simplify

d^2 + 1024 = d^2 + 8d + 16 + 576

1024 - 16 - 576  =  8d

432  = 8d

d =432 / 8   =  54

And

d^2 + 32^2  = r^2

54^2 + 32^2 = r^2

3940  = r^2

So  we have another right triangle  such that the distance from the center of the circle to WX = d + 2

So

(d + 2)^2  + ( WX / 2)^2  = r^2

(54 + 2)^2  +  WX^2 / 4  = 3940

56^2  + WX^2 / 4  =3940

3136 + WX^2 /  4  = 3940

WX^2 =  4 ( 3940 - 3136)

WX^2  = 4 * 804

WX^2  = 3216

WX = sqrt (3216)   ≈  56.7

This makes sense.....WX should  be > 48  but < 64

Simplify as

x^2 + nx + 316 < 0

Set as an equality 

We will have real solutions whenever the discriminant is  ≥ 0

So

n^2 - 4(1)(316) ≥ 0

n^2 - 1264 ≥ 0

n^2 ≥ 1264             take both roots

n ≥ sqrt (1264)  → ≈ 35.6   or    n  ≤  -sqrt (1264) → ≈  -35.6

Thus....we have real solutions when  n  = ( inf , ≈ -35.6] U [ ≈ 35.6 , inf )

So  the integers for n that  make this  have NO real solutions  in x are the integers between these two values

n = [ -35 , 35 ]  =    2(35) + 1 =   71  integers

55 *35 * 30  = 57750 cm^3

62 L  =  62000 cm^3

So

62000 - 57750  = 4250 cm^3  overflows  =  4.25 L  =  4L +  250 ml

https://web2.0calc.com/questions/rates_57