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Degger [83]
3 years ago
11

5) Payton runs a kiosk on Route 66. Half the canned drinks in stock were sold out and empty cans were sent for recycling. Payton

bought 68 more canned drinks to restock the kiosk. If there is a total of 200 cans now, how many canned drinks were in stock initially? ​
Mathematics
1 answer:
Zanzabum3 years ago
4 0

Answer: There were 264 cans initially stocked.

Step-by-step explanation:

c/2 + 68 = 200

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If Beth has 5 books and Tom has 3 how many do they have in all?
aalyn [17]

Answer:

8 books

Step-by-step explanation:

5+3=8

4 0
3 years ago
Read 2 more answers
How do I find the missing sides of this triangle by using trigonometry
Lemur [1.5K]

Step-by-step explanation:

Use SOH-CAH-TOA.

Sine = Opposite / Hypotenuse

Cosine = Adjacent / Hypotenuse

Tangent = Opposite / Adjacent

You're given an angle, the adjacent side to that angle, and the hypotenuse.  So use cosine.

Cosine = Adjacent / Hypotenuse

cos 37° = 48 / x

Multiply both sides by x.

x cos 37° = 48

Divide both sides by cos 37°.

x = 48 / cos 37°

If desired, use a calculator to evaluate.

x ≈ 60.1

8 0
3 years ago
Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75.percent f
irinina [24]

Answer:

33%

Step-by-step explanation:

Assuming the weight of the mixture to be 100g**, then the weight of ryegrass in the mixture would be 30g.

Also, assume the weight mixture X used in the mixture is Xg, then the weight of mixture Y used in the mixture would be (100-X)g.

So we can now equate the parts of the ryegrass in the mixture as:

0.4X + 0.25(100-X) = 30

<=> 0.4X + 25 - 0.25X = 30

<=> 0.15X = 5

<=> X = 5/0.15 = 500/15 = 100/3

So the weight of mixture X as a percentage of the weight of the mixture

= (weight of X/weight of mixture) * 100%

= (100/3)/100 * 100%

= 33%

3 0
3 years ago
Find the critical points of the function f(x, y) = 8y2x − 8yx2 + 9xy. Determine whether they are local minima, local maxima, or
NARA [144]

Answer:

Saddle point: (0,0)

Local minimum: (\frac{3}{8}, -\frac{3}{8})

Local maxima: (0,-\frac{9}{8}), (\frac{9}{8},0)

Step-by-step explanation:

The function is:

f(x,y) = 8\cdot y^{2}\cdot x -8\cdot y\cdot x^{2} + 9\cdot x \cdot y

The partial derivatives of the function are included below:

\frac{\partial f}{\partial x} = 8\cdot y^{2}-16\cdot y\cdot x+9\cdot y

\frac{\partial f}{\partial x} = y \cdot (8\cdot y -16\cdot x + 9)

\frac{\partial f}{\partial y} = 16\cdot y \cdot x - 8 \cdot x^{2} + 9\cdot x

\frac{\partial f}{\partial y} = x \cdot (16\cdot y - 8\cdot x + 9)

Local minima, local maxima and saddle points are determined by equalizing  both partial derivatives to zero.

y \cdot (8\cdot y -16\cdot x + 9) = 0

x \cdot (16\cdot y - 8\cdot x + 9) = 0

It is quite evident that one point is (0,0). Another point is found by solving the following system of linear equations:

\left \{ {{-16\cdot x + 8\cdot y=-9} \atop {-8\cdot x + 16\cdot y=-9}} \right.

The solution of the system is (3/8, -3/8).

Let assume that y = 0, the nonlinear system is reduced to a sole expression:

x\cdot (-8\cdot x + 9) = 0

Another solution is (9/8,0).

Now, let consider that x = 0, the nonlinear system is now reduced to this:

y\cdot (8\cdot y+9) = 0

Another solution is (0, -9/8).

The next step is to determine whether point is a local maximum, a local minimum or a saddle point. The second derivative test:

H = \frac{\partial^{2} f}{\partial x^{2}} \cdot \frac{\partial^{2} f}{\partial y^{2}} - \frac{\partial^{2} f}{\partial x \partial y}

The second derivatives of the function are:

\frac{\partial^{2} f}{\partial x^{2}} = 0

\frac{\partial^{2} f}{\partial y^{2}} = 0

\frac{\partial^{2} f}{\partial x \partial y} = 16\cdot y -16\cdot x + 9

Then, the expression is simplified to this and each point is tested:

H = -16\cdot y +16\cdot x -9

S1: (0,0)

H = -9 (Saddle Point)

S2: (3/8,-3/8)

H = 3 (Local maximum or minimum)

S3: (9/8, 0)

H = 9 (Local maximum or minimum)

S4: (0, - 9/8)

H = 9 (Local maximum or minimum)

Unfortunately, the second derivative test associated with the function does offer an effective method to distinguish between local maximum and local minimums. A more direct approach is used to make a fair classification:

S2: (3/8,-3/8)

f(\frac{3}{8} ,-\frac{3}{8} ) = - \frac{27}{64} (Local minimum)

S3: (9/8, 0)

f(\frac{9}{8},0) = 0 (Local maximum)

S4: (0, - 9/8)

f(0,-\frac{9}{8} ) = 0 (Local maximum)

Saddle point: (0,0)

Local minimum: (\frac{3}{8}, -\frac{3}{8})

Local maxima: (0,-\frac{9}{8}), (\frac{9}{8},0)

4 0
3 years ago
But whats the answer
Gekata [30.6K]

Question: Mia and Kai want to rent a canoe at the boathouse on the river. It costs $6.95 to rent a canoe for 1 hour. The price per hour stays the same for up to 3 hours of rental. After 3 hours, the cost decreases to $4.50 per hour. How much would it cost Mia and Kai to rent a canoe for 4 hours?

Answer:

$23.30

Step-by-step explanation:

To do this I did 6.45(3) + 6.45-4.50

20. 85 + 2.45

23.30

Hope this helps ya!

5 0
2 years ago
Read 2 more answers
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