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sergeinik [125]
3 years ago
13

The cost of a single ticket depends on the number of tickets, t, purchased in a group, as represented by the function C(t). C(t)

= StartLayout Enlarged left-brace 1st row 1st column dollar-sign 18.50 2nd column 1 less-than-or-equal-to t less-than 12 2nd row 1st column dollar-sign 16.00 2nd column 12 less-than-or-equal-to t less-than 20 3rd row 1st column dollar-sign 14.50 2nd column 20 less-than-or-equal-to t EndLayout What is the total cost for 20 tickets? $14.50 $16.00 $290.00 $320.00
Mathematics
2 answers:
Over [174]3 years ago
5 0

Answer: C. $290.00

Step-by-step explanation:

kobusy [5.1K]3 years ago
3 0

Answer:

The total cost for 20 tickets:

20 · 14.50 = $290.00

Answer: C ) $290.00 on edge nuity

Step-by-step explanation:

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3 years ago
MARSVECTORCALC6 3.4.020. My Notes A rectangular box with no top is to have a surface area of 64 m2. Find the dimensions (in m) t
ioda

Answer:

We would have

                                    l =w =\frac{8\sqrt{3}}{3} \\h = \frac{8\sqrt{3}}{6}

where " l " is  length, " w"  is width and "h" is height.

Step-by-step explanation:

Step 1

Remember that

         Surface area for a box with no top = lw+2lh+2wh = 64

where " l " is  length, " w"  is width and "h" is height.

 Step 2.

Remember as well that

                              Volume of the box = l*w*h

Step 3

 We can now use lagrange multipliers.  Lets say,

                                    F(l,w,h) = lwh

and

                                g(l,w,h) = lw+2lh+2wh = 64

By the lagrange multipliers method we know that                            

 

                                                     \nabla F  = \lambda \nabla g

Step 4

Remember that

                          \nabla F  = (wh,lh,lw)

and

                      \nabla g = (w+2h,l+2h , 2w+2l)

So basically you will have the system of equations

                              wh = \lambda (w+2h)\\lh = \lambda (l+2h)\\lw = \lambda (2w+2l)

Now, remember that you can multiply the first eqation, by "l" the second equation by "w" and the third one by "h" and you would get

                                   lwh = l\lambda (w+2h)\\\\lwh = w\lambda (l+2h)\\\\lwh = h\lambda (2w+2l)

Then you would get

                      l\lambda (w+2h) = w\lambda (l+2h) =  h\lambda (2w+2l)

You can get rid of \lambda from these equations and you would get

                         lw+2lh = lw+2wh =  2wh+2lh

And from those equations you would get

                                             l = w =2h

Now remember the original equation

                                    lw+2lh+2wh = 64

If we plug in what we just got, we would have

                                 l^{2} + l^{2} + l^2   =  64 \\3l^{2} = 64 \\l = w = \frac{8\sqrt{3} }{3} \\h = \frac{8\sqrt{3} }{6}

                       

                                                 

7 0
3 years ago
Which equation is equivalent to 4^x+3=64
Ainat [17]
The answer is B. Hope this helps!(:
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3 years ago
What's the prime factorization of 78​
Masja [62]

Answer:

2,3 &13

Step-by-step explanation:

78=2times39=3times13

therefore the prime factors of 78=2,3,13

4 0
3 years ago
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