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Leto [7]
3 years ago
6

THIS IS URGENT I NEED THIS AS SOON AS POSSIBLE! YOUR HELP WOULD BE GREATLY APPRECIATED!

Mathematics
2 answers:
serious [3.7K]3 years ago
7 0

Answer:

Strawberries sold 44 baskets

Raspberries sold 52 baskets

Step-by-step explanation:

4x + 9y = 644 \\ x + y = 96 \\ 4x + 4y = 384  \\ 5y = 260 \\ y = 52 \\ x = 44

dsp733 years ago
7 0

Answer: 44 baskets of strawberries were sold.

Step-by-step explanation:

Let x represent the number of basket of strawberries that were sold.

Let y represent the number of basket of raspberries that were sold.

In one morning, Katya sells 96. This means that

x + y = 96

At Katya’s fruit stand, a basket of strawberries costs $4 and a basket of raspberries costs $9. The total amount gotten from all the baskets sold is $644. It means that

4x + 9y = 644 - - - - - - - - - - - -1

Substituting x = 96 - y into equation 1, it becomes

4(96 - y) + 9y = 644

384 - 4y + 9y = 644

- 4y + 9y = 644 - 384

5y = 260

y = 260/5 = 52

x = 96 - y = 96 - 52

x = 44

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Leni [432]

The question is incomplete. Here is the complete question.

Find the measurements (the lenght L and the width W) of an inscribed rectangle under the line y = -\frac{3}{4}x + 3 with the 1st quadrant of the x & y coordinate system such that the area is maximum. Also, find that maximum area. To get full credit, you must draw the picture of the problem and label the length and the width in terms of x and y.

Answer: L = 1; W = 9/4; A = 2.25;

Step-by-step explanation: The rectangle is under a straight line. Area of a rectangle is given by A = L*W. To determine the maximum area:

A = x.y

A = x(-\frac{3}{4}.x + 3)

A = -\frac{3}{4}.x^{2}  + 3x

To maximize, we have to differentiate the equation:

\frac{dA}{dx} = \frac{d}{dx}(-\frac{3}{4}.x^{2}  + 3x)

\frac{dA}{dx} = -3x + 3

The critical point is:

\frac{dA}{dx} = 0

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Substituing:

y = -\frac{3}{4}x + 3

y = -\frac{3}{4}.1 + 3

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So, the measurements are x = L = 1 and y = W = 9/4

The maximum area is:

A = 1 . 9/4

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6 0
3 years ago
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We know that

two compound propositions are said to be logically equivalent if they have same corresponding truth values in the truth table.

The truth table is as follows :

P     Q      ∼P     ∼Q     P⇔ Q    ∼P ∨ Q     ∼Q ∨ P        (∼P ∨ Q)∧(∼Q ∨ P)

T     T         F        F             T            T                   T                       T

T     F         F        T             F             F                   T                       F

F     T         T        F             F            T                   F                       F

F     F         T        T             T            T                   T                       T

Since the corresponding truth vales for P ⇔ Q and (∼P ∨ Q)∧(∼Q ∨ P) are same, so the given propositions are logically equivalent.

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