Let the amount of salmon Sasha bought be s.
Let the amount of whitefish Sasha bought be w.
Let the amount of swordfish Sasha bought be s.
Let the amount of haddock Sasha bought be h.
We have the following information:
s + w + s + h = 34 (equation 1)
9.6s + 7.2w + 8.4s + 6.9h = 5.4 (equation 2)
Simplifying equation 2, we get:
18s + 7.2w + 6.9h = 5.4 (equation 3)
Subtracting equation 3 from equation 2, we get:
9.6s + 8.4s = 0
18s + 7.2w - 6.9h = 5.4
-------------
-8.4s = -8.4
s = 1
Plugging in the value of s into equation 1, we get:
1 + w + 1 + h = 34
2 + w + h = 34
w + h = 32 (equation 4)
Now, plugging in the values of s and w into equation 3, we get:
18 + 7.2w + 6.9h = 5.4
7.2w + 6.9h = -12.6
Multiplying equation 4 by -6.9 and adding it to the above equation, we get:
-6.9w - 6.9h = -221.8
7.2w + 6.9h = -12.6
-----------------
0.3w = -234.4
w = -234.4 ÷ 0.3
w = -781.33333 (rounded to 6 decimal places)
Substituting the value of w into equation 4, we get:
-781.33333 + h = 32
h = 32 + 781.33333
h = 813.33333 (rounded to 6 decimal places)
So Sasha bought 1 pound of salmon, -781.33333 (rounded to 6 decimal places) pounds of whitefish, and 813.33333 (rounded to 6 decimal places) pounds of haddock.
At the fish counter, salmon was selling for $9.60 a pound, whitefish for $7.20 a pound, swordfish for $8.40 a pound, and haddock for $6.90 a pound. Sasha bought 34
lb. of fish for $5.40. What fish did he choose?
8 answers
wrong
Apologies for the mistake. Let's solve the problem correctly:
Let x be the amount of salmon Sasha bought.
Let y be the amount of whitefish Sasha bought.
Let z be the amount of swordfish Sasha bought.
Let w be the amount of haddock Sasha bought.
We have the following information:
x + y + z + w = 34 (equation 1) [since Sasha bought a total of 34 pounds of fish]
9.6x + 7.2y + 8.4z + 6.9w = 5.4 (equation 2) [since Sasha paid $5.40 for the fish]
In this problem, we are looking for the combination of fish Sasha bought. So we need to solve this system of equations.
Multiplying equation 1 by 9.6, we get:
9.6x + 9.6y + 9.6z + 9.6w = 326.4 (equation 3)
Subtracting equation 3 from equation 2, we get:
9.6x + 7.2y + 8.4z + 6.9w - (9.6x + 9.6y + 9.6z + 9.6w) = 5.4 - 326.4
-2.4y - 1.2z - 2.7w = -321 (equation 4)
Now, let's solve the system of equations.
From equation 4, we can rewrite it as:
2.4y + 1.2z + 2.7w = 321 (equation 5) [by multiplying both sides of the equation by -1]
Now, we can use a substitution method.
From equation 1, we can express x in terms of y, z, and w by subtracting the other variables from 34:
x = 34 - y - z - w
Substituting this into equation 5, we get:
2.4y + 1.2z + 2.7w = 321 (equation 5)
By multiplying equation 5 by 10 to get rid of the decimals, we have:
24y + 12z + 27w = 3210 (equation 6)
Now, let's solve equations 5 and 6 using the substitution method:
We can rewrite equation 5 as:
w = (321 - 2.4y - 1.2z) / 2.7 (equation 5') [by isolating w]
Substituting equation 5' into equation 6, we get:
24y + 12z + 27((321 - 2.4y - 1.2z) / 2.7) = 3210
Simplifying the equation yields:
24y + 12z + 9(321 - 2.4y - 1.2z) = 3210
Distributing 9 to each term inside the parentheses:
24y + 12z + 2898 - 21.6y - 10.8z = 3210
Combine like terms:
12.4y + 1.2z = 312
Now, we have a linear equation with two variables. We can solve it by isolating one variable and substituting it back into the equation.
Let's solve for z in terms of y:
z = 312 - 12.4y / 1.2 (equation 7)
Since z represents the amount of fish Sasha bought, it must be a whole number. Therefore, we will test different integer values for y to find a corresponding integer value for z.
If we let y = 0, substituting into equation 7:
z = 312 - 12.4(0) / 1.2
z = 312
This is possible since z can represent the amount of any fish that Sasha bought.
So, if Sasha chose 0 pounds of whitefish, the answer is: Salmon.
If we let y = 1, substituting into equation 7:
z = 312 - 12.4(1) / 1.2
z = 312 - 12.4 / 1.2
z ≈ 302.333333
This is not a whole number, so it's not a valid solution.
If we let y = 2, substituting into equation 7:
z = 312 - 12.4(2) / 1.2
z = 312 - 24.8 / 1.2
z ≈ 292.666667
This is not a whole number either.
If we let y = 3, substituting into equation 7:
z = 312 - 12.4(3) / 1.2
z = 312 - 37.2 / 1.2
z ≈ 283
This is a whole number, so a valid solution is: Salmon, 3 pounds of Whitefish.
If we let y = 4, substituting into equation 7:
z = 312 - 12.4(4) / 1.2
z = 312 - 49.6 / 1.2
z ≈ 273.333333
This is not a whole number either.
If we let y = 5, substituting into equation 7:
z = 312 - 12.4(5) / 1.2
z = 312 - 62 / 1.2
z ≈ 263
This is a whole number, so another valid solution is: Salmon, 5 pounds of Whitefish.
Note: There may be more valid combinations, but these are two possibilities.
Therefore, Sasha either bought Salmon or Salmon and 3 or 5 pounds of Whitefish.
Let x be the amount of salmon Sasha bought.
Let y be the amount of whitefish Sasha bought.
Let z be the amount of swordfish Sasha bought.
Let w be the amount of haddock Sasha bought.
We have the following information:
x + y + z + w = 34 (equation 1) [since Sasha bought a total of 34 pounds of fish]
9.6x + 7.2y + 8.4z + 6.9w = 5.4 (equation 2) [since Sasha paid $5.40 for the fish]
In this problem, we are looking for the combination of fish Sasha bought. So we need to solve this system of equations.
Multiplying equation 1 by 9.6, we get:
9.6x + 9.6y + 9.6z + 9.6w = 326.4 (equation 3)
Subtracting equation 3 from equation 2, we get:
9.6x + 7.2y + 8.4z + 6.9w - (9.6x + 9.6y + 9.6z + 9.6w) = 5.4 - 326.4
-2.4y - 1.2z - 2.7w = -321 (equation 4)
Now, let's solve the system of equations.
From equation 4, we can rewrite it as:
2.4y + 1.2z + 2.7w = 321 (equation 5) [by multiplying both sides of the equation by -1]
Now, we can use a substitution method.
From equation 1, we can express x in terms of y, z, and w by subtracting the other variables from 34:
x = 34 - y - z - w
Substituting this into equation 5, we get:
2.4y + 1.2z + 2.7w = 321 (equation 5)
By multiplying equation 5 by 10 to get rid of the decimals, we have:
24y + 12z + 27w = 3210 (equation 6)
Now, let's solve equations 5 and 6 using the substitution method:
We can rewrite equation 5 as:
w = (321 - 2.4y - 1.2z) / 2.7 (equation 5') [by isolating w]
Substituting equation 5' into equation 6, we get:
24y + 12z + 27((321 - 2.4y - 1.2z) / 2.7) = 3210
Simplifying the equation yields:
24y + 12z + 9(321 - 2.4y - 1.2z) = 3210
Distributing 9 to each term inside the parentheses:
24y + 12z + 2898 - 21.6y - 10.8z = 3210
Combine like terms:
12.4y + 1.2z = 312
Now, we have a linear equation with two variables. We can solve it by isolating one variable and substituting it back into the equation.
Let's solve for z in terms of y:
z = 312 - 12.4y / 1.2 (equation 7)
Since z represents the amount of fish Sasha bought, it must be a whole number. Therefore, we will test different integer values for y to find a corresponding integer value for z.
If we let y = 0, substituting into equation 7:
z = 312 - 12.4(0) / 1.2
z = 312
This is possible since z can represent the amount of any fish that Sasha bought.
So, if Sasha chose 0 pounds of whitefish, the answer is: Salmon.
If we let y = 1, substituting into equation 7:
z = 312 - 12.4(1) / 1.2
z = 312 - 12.4 / 1.2
z ≈ 302.333333
This is not a whole number, so it's not a valid solution.
If we let y = 2, substituting into equation 7:
z = 312 - 12.4(2) / 1.2
z = 312 - 24.8 / 1.2
z ≈ 292.666667
This is not a whole number either.
If we let y = 3, substituting into equation 7:
z = 312 - 12.4(3) / 1.2
z = 312 - 37.2 / 1.2
z ≈ 283
This is a whole number, so a valid solution is: Salmon, 3 pounds of Whitefish.
If we let y = 4, substituting into equation 7:
z = 312 - 12.4(4) / 1.2
z = 312 - 49.6 / 1.2
z ≈ 273.333333
This is not a whole number either.
If we let y = 5, substituting into equation 7:
z = 312 - 12.4(5) / 1.2
z = 312 - 62 / 1.2
z ≈ 263
This is a whole number, so another valid solution is: Salmon, 5 pounds of Whitefish.
Note: There may be more valid combinations, but these are two possibilities.
Therefore, Sasha either bought Salmon or Salmon and 3 or 5 pounds of Whitefish.
...so what's the answer?
Apologies for the confusion. Based on the given information, we can determine the fish Sasha chose.
From the information provided, Sasha bought a total of 34 pounds of fish for $5.40.
Let x be the amount of salmon Sasha bought.
Let y be the amount of whitefish Sasha bought.
Let z be the amount of swordfish Sasha bought.
Let w be the amount of haddock Sasha bought.
We have the following equations:
x + y + z + w = 34 (equation 1) [since Sasha bought a total of 34 pounds of fish]
9.6x + 7.2y + 8.4z + 6.9w = 5.4 (equation 2) [since Sasha paid $5.40 for the fish]
To solve this system of equations, we can use a method called substitution.
From equation 1, we can express x in terms of y, z, and w by subtracting the other variables from 34:
x = 34 - y - z - w
Substituting this expression for x into equation 2, we get:
9.6(34 - y - z - w) + 7.2y + 8.4z + 6.9w = 5.4
Expanding and simplifying the equation, we have:
326.4 - 9.6y - 9.6z - 9.6w + 7.2y + 8.4z + 6.9w = 5.4
Combining like terms:
320.4 - 2.4y - 1.2z - 2.7w = 5.4
Subtracting 320.4 from both sides:
-2.4y - 1.2z - 2.7w = -315
Now we have a linear equation in three variables. To find a specific solution, we need additional information.
If there is no further information, then we cannot determine the exact combination of fish Sasha chose using this given information alone. We need additional information, such as the price of each type of fish or the exact amount Sasha paid for each type, to find the specific combination of fish Sasha bought.
From the information provided, Sasha bought a total of 34 pounds of fish for $5.40.
Let x be the amount of salmon Sasha bought.
Let y be the amount of whitefish Sasha bought.
Let z be the amount of swordfish Sasha bought.
Let w be the amount of haddock Sasha bought.
We have the following equations:
x + y + z + w = 34 (equation 1) [since Sasha bought a total of 34 pounds of fish]
9.6x + 7.2y + 8.4z + 6.9w = 5.4 (equation 2) [since Sasha paid $5.40 for the fish]
To solve this system of equations, we can use a method called substitution.
From equation 1, we can express x in terms of y, z, and w by subtracting the other variables from 34:
x = 34 - y - z - w
Substituting this expression for x into equation 2, we get:
9.6(34 - y - z - w) + 7.2y + 8.4z + 6.9w = 5.4
Expanding and simplifying the equation, we have:
326.4 - 9.6y - 9.6z - 9.6w + 7.2y + 8.4z + 6.9w = 5.4
Combining like terms:
320.4 - 2.4y - 1.2z - 2.7w = 5.4
Subtracting 320.4 from both sides:
-2.4y - 1.2z - 2.7w = -315
Now we have a linear equation in three variables. To find a specific solution, we need additional information.
If there is no further information, then we cannot determine the exact combination of fish Sasha chose using this given information alone. We need additional information, such as the price of each type of fish or the exact amount Sasha paid for each type, to find the specific combination of fish Sasha bought.
Hmmph. Here's all the answers.
A1: Negative
A2: 1,037
A3: -4 1/4
A4: -0.38
A5: whitefish
A6: -75.00
A7: always negative
A8: 52.28
A9: 4 3/8
A10: 4/5
A11: 0.4
A12: 6.35
A13: -$4.30
A14: The Commutative Property of Multiplication says that changing the order in which you multiply the factors does not change the answer. a x b = b x a
A15: 10
A1: Negative
A2: 1,037
A3: -4 1/4
A4: -0.38
A5: whitefish
A6: -75.00
A7: always negative
A8: 52.28
A9: 4 3/8
A10: 4/5
A11: 0.4
A12: 6.35
A13: -$4.30
A14: The Commutative Property of Multiplication says that changing the order in which you multiply the factors does not change the answer. a x b = b x a
A15: 10
I apologize for the confusion, but the information you provided does not seem to be related to the original problem regarding Sasha buying fish. If you have any further questions or need assistance with a different topic, please let me know and I'll be happy to help.
THANK YOU BIRD BRAIN I WANT TO MAKE OUT WITH YOU