A quantity Q is obtained by adding three quantities.

Question

The first is a constant, the second varies directly with the square root of y and the third varies directly with
the cube root of y. When y = 1, Q = 60. When
y = 64, Q = 230 and when y = 729, Q = 660
Find the constant.

Answer 1

Let the constant be a.
The second quantity varies directly with the square root of y, so let it be bx.
The third quantity varies directly with the cube root of y, so let it be cy^(1/3).
Thus, Q = a + bx + cy^(1/3).

When y = 1, Q = 60:
60 = a + b + c

When y = 64, Q = 230:
230 = a + 8b + 4c

When y = 729, Q = 660:
660 = a + 27b + 9c

Solve the system of equations:
a + b + c = 60
a + 8b + 4c = 230
a + 27b + 9c = 660

Subtract equation 1 from equation 2 to eliminate a:
a + 8b + 4c - (a + b + c) = 230 - 60
7b + 3c = 170

Subtract equation 1 from equation 3 to eliminate a:
a + 27b + 9c - (a + b + c) = 660 - 60
26b + 8c = 600

Multiply equation 3 by 7 and equation 2 by 26, and subtract to eliminate b:
26(7b + 3c) - 7(26b + 8c) = 26(170) - 7(600)
98c = 520
c = 520/98
c = 5.306

Substitute c = 5.306 into equation 1:
a + b + 5.306 = 60
a + b = 60 - 5.306
a + b = 54.694

Substitute c = 5.306 into equation 2:
a + 8b + 4(5.306) = 230
a + 8b + 21.224 = 230
a + 8b = 230 - 21.224
a + 8b = 208.776

Substitute c = 5.306 into equation 3:
a + 27b + 9(5.306) = 660
a + 27b + 47.754 = 660
a + 27b = 660 - 47.754
a + 27b = 612.246

Subtract 8 times equation 4 from equation 3 to eliminate a and b:
a + 27b - (8a + 8b) = 612.246 - 8(54.694)
-7a + 19b = 612.246 - 437.552
-7a + 19b = 174.694

Multiply equation 5 by 7 and add to equation 6 to eliminate b:
-49a + 133b + 7(-7a + 19b) = 1212.462 + 1222.858
-49a + 133b - 49a + 133b = 2435.32

Combine like terms:
84b = 2435.32
b = 2435.32/84
b = 29.040

Substitute b = 29.040 into equation 4:
a + (29.040) + 5.306 = 54.694
a + 34.346 = 54.694
a = 54.694 - 34.346
a = 20.348

So, the constant is approximately 20.348.

Answer 2

huh? Where did x come from? clearly,
q = a + b√y + c∛y
your solution went astray, since it should be.
so (a,b,c) = (30,20,10)
so the constant is a=30

Answer 3

Apologies for the confusion in my previous response.

Given:
Q = a + b√y + c∛y

When y = 1, Q = 60:
60 = a + b + c

When y = 64, Q = 230:
230 = a + 8b + 16c

When y = 729, Q = 660:
660 = a + 27b + 81c

Solve the system of equations:
a + b + c = 60
a + 8b + 16c = 230
a + 27b + 81c = 660

We can solve this system of equations using various methods such as substitution or elimination. Here, I'll use substitution:

From equation 1, we have: a = 60 - b - c

Substitute this value of a into equations 2 and 3:
(60 - b - c) + 8b + 16c = 230
(60 - b - c) + 27b + 81c = 660

Simplify equation 2:
60 - b - c + 8b + 16c = 230
-b + 7b + 15c = 170
6b + 15c = 170

Simplify equation 3:
60 - b - c + 27b + 81c = 660
-b + 26b + 80c = 600
25b + 80c = 600

Now we have two equations in two variables:
6b + 15c = 170
25b + 80c = 600

Multiply equation 1 by 5:
30b + 75c = 850

Subtract this equation from equation 2 to eliminate b:
(25b + 80c) - (30b + 75c) = 600 - 850
-5b + 5c = -250
b - c = 50

Now we have two equations in two variables:
b - c = 50
6b + 15c = 170

Multiply equation 1 by 6:
6b - 6c = 300

Add this equation to equation 2 to eliminate c:
(6b + 15c) + (6b - 6c) = 170 + 300
12b + 9c = 470

Multiply equation 1 by 9:
9b - 9c = 450

Add this equation to equation 2 to eliminate c:
(12b + 9c) + (9b - 9c) = 470 + 450
21b = 920
b = 920/21
b ≈ 43.81

Substitute the value of b into equation 1:
43.81 - c = 50
c = 43.81 - 50
c ≈ -6.19

Substitute the values of b and c into equation 1:
a + 43.81 - 6.19 = 60
a + 37.62 = 60
a = 60 - 37.62
a ≈ 22.38

Therefore, the constant is approximately a = 22.38.