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viktelen [127]
3 years ago
11

A florist is filling a large order for a client. The client wants no more than 300 roses in vases. The smaller vase will contain

8 roses and the larger vase will contain 12 roses. The client requires that there are at least twice as many small vases as large vases. The client requires that there are at least 6 small vases and no more than 12 large vases.
Let x represent the number of small vases and y represent the number of large vases.



What constraints are placed on the variables in this situation?
Mathematics
1 answer:
d1i1m1o1n [39]3 years ago
4 0

Answer:

8x+12y\leq 300

x\geq 6

y\leq 12

x\geq 2y

Step-by-step explanation:

We are given that

x=Number of small vases

y=Number of large vases

Total number of roses not more than 300 in vases.

Number of roses in small vase atleast=8

Number of roses in large vase not more than =12

We have to find the constraints are placed on the variables in the given situation.

According to question

8x+12y\leq 300

x\geq 6

y\leq 12

x\geq 2y

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lora16 [44]

Answer:

f(3) = - 18

Step-by-step explanation:

To evaluate f(3) , substitute x = 3 into f(x) , that is

f(3) = - 4(3 + 1) - 2 = - 4(4) - 2 = - 16 - 2 = - 18

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Can someone literally cash app me a dollar at $AlphaDawggg it will really make my day
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If (5^2)p = 5^12, what is the value of p?
mr Goodwill [35]
Divide both sides by 5^2 (which is 25). So p = (5^12)/25. I can't be bothered to plug in into a calculator but be my guest.
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3 years ago
Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into
Alex_Xolod [135]

Answer:

a) the length of the wire for the circle = (\frac{60\pi }{\pi+4}) in

b)the length of the wire for the square = (\frac{240}{\pi+4}) in

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

         b=\frac{y}{4}

Area of the square which is L² can now be said to be;

A_S=(\frac{y}{4})^2 = \frac{y^2}{16}

On the otherhand; let the radius (r) of the  circle be;

2πr = 60-y

r = \frac{60-y}{2\pi }

Area of the circle which is πr² can now be;

A_C= \pi (\frac{60-y}{2\pi } )^2

     =( \frac{60-y}{4\pi } )^2

Total Area (A);

A = A_S+A_C

   = \frac{y^2}{16} +(\frac{60-y}{4\pi } )^2

For the smallest possible area; \frac{dA}{dy}=0

∴ \frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0

If we divide through with (2) and each entity move to the opposite side; we have:

\frac{y}{18}=\frac{(60-y)}{2\pi}

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

y= \frac{240}{\pi+4}

∴ since y= \frac{240}{\pi+4}, we can determine for the length of the circle ;

60-y can now be;

= 60-\frac{240}{\pi+4}

= \frac{(\pi+4)*60-240}{\pi+40}

= \frac{60\pi+240-240}{\pi+4}

= (\frac{60\pi}{\pi+4})in

also, the length of wire for the square  (y) ; y= (\frac{240}{\pi+4})in

The smallest possible area (A) = \frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

4 0
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Kelvin is making a salad. He can pick between spinach, lettuce, or arugula as the base. He can make it with or without tomatoes,
Sunny_sXe [5.5K]
He can make 27 different salads. Since each individual option has three different salads, and there are 9 options.

Basically multiply the number of total toppings by the number of toppings in each row

Answer: 27
6 0
3 years ago
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