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erastovalidia [21]
3 years ago
14

Strawberries are $2.50 A pound and cantaloupes are $2.25 at the local supermarket. Sally bought 7 pounds of the two kinds of fru

it for a family breakfast. If she spent exactly $16.75 and bought at least 1 pound of each fruit how many pounds of fruit did she buy there is no sales tax
Mathematics
1 answer:
Vlad1618 [11]3 years ago
7 0

Answer:

Sally bought 4 pounds of Strawberries and 3 pounds of Cantaloupes.

Step-by-step explanation:

Let the number of pounds of Strawberries bought by Sally = x

Let the number of pounds of Cantaloupes bought by Sally = y

\[x + y = 7\]  ---------------------------------(1)

Moreover,

\[2.5 x + 2.25 y = 16.75\] ---------------(2)

Solving (1) and (2) by substitution:

\[x = 7 - y\]

=> \[2.5 *(7-y) + 2.25 y = 16.75\]

=> \[17.5 - 2.5y + 2.25 y = 16.75\]

=> \[0.25y = 0.75\]

=> \[y = 3\]

From (1), x = 7-3 = 4

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Answer:

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If 50 apples cost $25 to find the cost of 1 apple, we divide the total cost by the total number of items. In this case, we do 25/50.

One apple costs 0.50. If we need 75 apples then we simply multiply 0.50 by 75 which gets us 37.50.

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lim n → ∞ an = ∞ means that as n approaches infinity (becomes large), an approaches infinity (becomes large).

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2 years ago
<img src="https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B4%7D%20%20-%206%20%7Bx%7D%5E%7B3%7D%20%20%2B%2022%20%7Bx%7D%5E%7B2%7D%20%20-%2
alexira [117]

Answer:

x = 2, 1 + 3i, 1 − 3i

Step-by-step explanation:

Find the Roots (Zeros)

x^4 − 6x^3 + 22x^2 − 48x + 40

Set x^4 − 6x^3 + 22x^2 − 48x + 40 equal to 0. x^4 − 6x^3 + 22x^2 − 48x + 40 = 0

Solve for x.

Factor the left side of the equation.

Factor x^4 − 6x^3 + 22x^2 − 48x + 40 using the rational roots test.

(x − 2) (x^3 − 4x^2 + 14x − 20) = 0

 Factor x^3 − 4x^2 + 14x − 20 using the rational roots test.

(x − 2) (x − 2) (x2 − 2x + 10) = 0

 Combine like factors.

(x − 2)2 (x^2 − 2x + 10) = 0

If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.

(x − 2)^2 = 0

x^2 − 2x + 10 = 0

 Set (x − 2)^2 equal to 0 and solve for x.

Set (x − 2)^2 equal to 0.

 (x − 2)^2 = 0

Solve (x − 2)^2 = 0 for x.

x = 2

 Set x^2 − 2x + 10 equal to 0 and solve for x.

Set x^2 − 2x + 10 equal to 0. x^2 − 2x + 10 = 0

Solve x^2 − 2x + 10 = 0 for x.

Use the quadratic formula to find the solutions.

−b ± (√b^2 − 4 (ac) )/2a

Substitute the values a = 1, b = −2, and c = 10 into the quadratic formula and solve for x.

2 ± (√(−2)^2 − 4 ⋅ (1 ⋅ 10))/2 ⋅ 1

Simplify.

Simplify the numerator.

  x =    2 ± 6i/ 2.1

Multiply 2 by 1

 x =  2 ± 6i/2⋅1

 Simplify

  2 ± 6i/2  

   x = 1 ± 3i

The final answer is the combination of both solutions.

x = 1 + 3i, 1 − 3i

The final solution is all the values that make (x − 2)2 (x2 − 2x + 10) = 0 true.

x = 2, 1 + 3i, 1 − 3i

3 0
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