A quantity Q is obtained by adding three quantities.
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.
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
To find the constant, we need to analyze the given information and determine the relationships between the quantities.
Let's denote the constant as C, the first quantity as A, the second quantity as B, and the third quantity as D.
According to the problem, we have:
Q = A + B + D
We are also given that:
When y = 1, Q = 60
When y = 64, Q = 230
When y = 729, Q = 660
We can now set up three equations using the given information:
Equation 1: 60 = A + B + D (when y = 1)
Equation 2: 230 = A + 8B + 9D (when y = 64, as the square root of 64 is 8 and the cube root of 64 is 4)
Equation 3: 660 = A + 12B + 27D (when y = 729, as the square root of 729 is 27 and the cube root of 729 is 9)
Now we have a system of three equations and three unknowns (A, B, D). We can solve this system of equations by elimination or substitution to find the values of A, B, and D. Once we have the values, we can determine the constant C by setting y = 0 (or any other suitable value) and solving for C using the equation Q = A + B + D.
Let's solve the system of equations to find the values of A, B, and D:
Equation 1: 60 = A + B + D .........(1)
Equation 2: 230 = A + 8B + 9D .........(2)
Equation 3: 660 = A + 12B + 27D .......(3)
Subtracting equation (1) from equation (2), we get:
170 = 7B + 8D .........(4)
Subtracting equation (1) from equation (3), we get:
600 = 11B + 26D ........(5)
Now we have two equations (equations 4 and 5) with two unknowns (B and D). We can solve this system of equations using substitution or elimination.
Multiplying equation (4) by 11 and equation (5) by 7 to eliminate B, we get:
1870 = 77B + 88D
4200 = 77B + 182D
Subtracting equation (4) from equation (5), we get:
4200 - 1870 = 182D - 88D
2330 = 94D
Dividing both sides by 94, we get:
D = 2330/94
D = 24.7872 (rounded to four decimal places)
Now, substituting the value of D back into equation (4), we get:
170 = 7B + 8(24.7872)
170 = 7B + 198.2976
-28.2976 = 7B
B = -4.0425 (rounded to four decimal places)
Substituting the values of B and D into equation (1), we get:
60 = A - 4.0425 + 24.7872
60 = A + 20.7447
A = 60 - 20.7447
A = 39.2553 (rounded to four decimal places)
Now we have the values of A, B, and D:
A = 39.2553
B = -4.0425
D = 24.7872
Finally, to find the constant C, we set y = 0 and solve for C using the equation Q = A + B + D:
C = Q - (B + D)
C = 60 - (-4.0425 + 24.7872)
C = 60 + 4.0425 - 24.7872
C = 39.2553
Hence, the constant is C ≈ 39.2553.
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.
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.