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2 + 3 + 1 = 6 total ratio

[3 /6] x 9 =4.5 pounds of walnuts in the mixture.

B/R = 5/4.............................(1)
(B + 4)/(R + 4) = 6/5............(2)

From (1) above:
4B = 5R
B = 5R/4  sub this into (2) above

(5R/4 + 4)/(R + 4) =6/5, solve for R
R = 16 - original number of red marbles            
B = 20 - original number of blue marbles.

Thanks for the compliment, Guest! I always strive to make my responses thorough while maintaining its comprehensibility.

I would not apologize. Your intent was clear: to serve the community well. You accomplished that goal, and you should receive an equal amount of commendment. It is also sometimes nice to view different approaches to the same problem.

Sorry X^2 !. I didn't see you working on it. Otherwise, I wouldn't have touched it, knowing how thorough you are.

Hey guest,

Let's try and solve for the question mark. Note that I have replaced the question mark with a more standard symbol, x. Below is a representation of the equation in LaTeX form:

\(1\frac{3}{5}(1999*\frac{1}{24}+x)-3\frac{5}{11}*3\frac{2}{3}=127\)

Solve for M:
(8 (M + 1999/24))/5 - 38/3 = 127

Put each term in M + 1999/24 over the common denominator 24: M + 1999/24 = (24 M)/24 + 1999/24:
8/5 (24 M)/24 + 1999/24 - 38/3 = 127

(24 M)/24 + 1999/24 = (24 M + 1999)/24:
8/5 (24 M + 1999)/24 - 38/3 = 127

8/24 = 8/(8×3) = 1/3:
(24 M + 1999)/(3×5) - 38/3 = 127

3×5 = 15:
(24 M + 1999)/15 - 38/3 = 127

Put each term in (24 M + 1999)/15 - 38/3 over the common denominator 15: (24 M + 1999)/15 - 38/3 = (24 M + 1999)/15 - 190/15:
(24 M + 1999)/15 - 190/15 = 127

(24 M + 1999)/15 - 190/15 = ((24 M + 1999) - 190)/15:
(24 M - 190 + 1999)/15 = 127

Add like terms. 1999 - 190 = 1809:
(24 M + 1809)/15 = 127

Multiply both sides of (24 M + 1809)/15 = 127 by 15:
(15 (24 M + 1809))/15 = 15×127

(15 (24 M + 1809))/15 = 15/15×(24 M + 1809) = 24 M + 1809:
24 M + 1809 = 15×127

15×127 = 1905:
24 M + 1809 = 1905

Subtract 1809 from both sides:
24 M + (1809 - 1809) = 1905 - 1809

1809 - 1809 = 0:
24 M = 1905 - 1809

1905 - 1809 = 96:
24 M = 96

Divide both sides of 24 M = 96 by 24:
(24 M)/24 = 96/24

24/24 = 1:
M = 96/24

The gcd of 96 and 24 is 24, so 96/24 = (24×4)/(24×1) = 24/24×4 = 4:
M = 4

Over the last 3 years salesperson A has earned twice as much as salesperson B, who in turned averaged 50% more then salesperson C. The three combined to make $220 000 over the three years.  How much did salesperson B earn on average in any year?

\(a=2b\\ \color{blue}b=\frac{a}{2}\\ b=1.5c\\ c=\frac{b}{1.5}=\frac{\frac{a}{2}}{1.5}\\ \color{blue}c=\frac{a}{3}\)

\((a+b+c)\cdot 3\ years=220000\$\\ (a+\frac{a}{2}+\frac{a}{3})\cdot 3\ years=220000\$\\ \frac{6a+3a+2a}{2\cdot3}=\frac{220000\$}{3\ years}\\ 11a=\frac{6\cdot 220000\$}{3\ years}\\ a=\frac{6\cdot 220000\$}{11\cdot 3\cdot years}\)

\(a=40000\ \$/year\)

\(b=\frac{a}{2}=\frac{40000}{2}\cdot \frac{\$}{year}\)

\(b=20000\ \$\ /\ year\)

Salesperson B earns 20 000 $ / year.

  !

Let salesperson C's earnings =C
Salesperson B's earnings =1.5C
Salesperson A's earnings =3C
C + 1.5C + 3C = $220,000
5.5C = $220,000
C = $220,000 / 5.5
C = $40,000 - total earnings of salesperson C over 3 years.
$40,000 x 1.5 = $60,000 - total earnings of salespeson B over 3 years.
$60,000 / 3 = $20,000 - Salesperson B's average earnings per year for 3 years.

Where are the numbers in your question??

The number you have listed, 11011_b can be represented in ANY base from 2 and up. When you say "is multiplied by b-1", I'm assuming you mean: (11011_b) - 1. If I'm wrong in my assumptions, let us know.

11011*11010+1001 =1011000111.00 - this is in base 2
                                =122002111.00 - this is in base 3
                                =121232111.00 - this is in base 4, 5, 6, 7, 8, 9,.........etc.

Note: You may choose the base you want, or is intended.

I think all your answers are correct! 👍

I got the same answer for each one:

k(m)  =  m² - 7m - 18

k(m)  =  (m - 9)(m + 2)

The zeros are the values of  m  that make  k(m)  equal zero, so the zeros are  9  and  -2 .

C)  One positive zero and one negative zero, where the positive zero has the larger absolute value

f(x)  =  2x² - 11x + 12

f(x)  =  2x² - 8x - 3x + 12

f(x)  =  2x(x - 4) - 3(x - 4)

f(x)  =  (x - 4)(2x - 3)

The zeros are  4  and  3/2  .

D)  Two positive zeros

s(r)  =  5r² + 49r - 10

s(r)  =  5r² + 50r - 1r - 10

s(r)  =  5r(r + 10) - 1(r + 10)

s(r)  =  (r + 10)(5r - 1)

The zeros are  -10  and  1/5 .

F)  One positive zero and one negative zero, where the negative zero has the larger absolute value

z(w)  =  w² + 7w + 6

z(w)  =  (w + 6)(w + 1)

The zeros are  -6  and  -1 .

E)  Two negative zeros

For the third question...

∠A  and  ∠B  must be acute angles because the third angle is 90°, and the sum of the angle measures in a triangle is 180°, and each angle has to have a measure greater than 0°. (If either  ∠A  or  ∠B  were  90°  then two of the sides would be parallel, and they could never meet.)

First let's find the sine and cosine of angle A .

sin A  =  opposite / hypotenuse

sin A  =  12 / 20

sin A  =  3 / 5

cos A  =  adjacent / hypotenuse

cos A  =  16 / 20

cos A  =  4 / 5

Next let's find the sine and cosine of angle B .

sin B  =  opposite / hypotenuse

sin B  =  16 / 20

sin B  =  4 / 5

cos B  =  adjacent / hypotenuse

cos B  =  12 / 20

cos B  =  3 / 5

S  =  2r³t

                       To solve this equation for  t ,  first let's divide both sides by  2 .

\(\frac{S}{2}\)  =  r³t

                       Next divide both sides by  r³ .

\(\frac{S}{2r^3}\)  =  t

t  =  \(\frac{S}{2r^3}\)

---------------------------

4x² - 376  =  24

                                 To solve this equation for  x , first let's add  376  to both sides.

4x²  =  24 + 376

4x²  =  400

                                 Then divide both sides by  4 .

x²  =  400/4

x²  =  100

                                 Then take the ± square root of both sides.

x  =  ±√100

x  =  ± 10

This means there are two values of  x  that make the equation true,

and those two values are 10  and  -10 .

I can help with part 2. 

angle A is cos(16/20) and angle B is sin(12/20)

OK......  

The only high school student I’ve (knowingly) trolled is Max Wong

Thank you! I appreciate your help!

BC  is a chord on the circle which has the same measure as the radial segments,  AB  and AC....thus  ABC  is equilateral with sides = 7

The measure of each angle  = 60°  ...so  angle A  = 60°

Hey...that's what we're here for...to help you  [if possible ]....!!!!

First one  : We know an opposite side to B  = 35   and the hypotenuse = 37

So....sin B  = 35 / 37

And we can use the  sine inverse  [arcsin ]  to  find B itself...so we have

arcsin (35/37 ] = B  ≈  71,07°

Second one

-1x^2 + 9x  = 6     subtract  6 from  both sides

-1x^2 + 9x  - 6 =  0    .....  a  =-1   b  = 9    c  = -6

So...the quad  formula  is     

-b  ± √ [ b^2  - 4ac ]                     -9 ± √ [ 9^2  - 4 (-1)(-6)]         -9 ± √ [ 81  - 24]

_______________     =             ___________________   =      ____________   =

       2a                                                    2(-1)                                   -2

-9 ± √ 57                 -9         √ 57               9            √ 57

_______   =            __  ±     ___   =         ___     ±   ___     

    -2                        -2            2                 2               2

Last one

Sign  in shape  of a rhombus with a 60° angle and sides of 7 cm

The area  =   7^2 * sin (60°)  ≈  49 * √3/ 2  =  (49/2)√​3  cm^2 = 42.4 cm^2

that's strange, I don't know why it's not showing up.

What is the measure of angle A, the central angle?

What is the measure of

Given  : ΔDEF ≅ ΔDGF

Congruent parts of congruent triangles are also congruent...so....

EF ≅  GF

Glad to  help....where I can....

 ∠DEF ≅ ∠DGF

The measure of what  ???