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PSYCHO15rus [73]
3 years ago
7

Can someone help me with this problem?

Mathematics
2 answers:
pashok25 [27]3 years ago
7 0
No, it is not a solution.

Let's plug 5 into the equation where p is.

10(5) - 2 > 97

50 - 2 > 97

48 > 97

This is false, 48 is less than 97.
Vlada [557]3 years ago
7 0
No because 10 x 5=50 then minus 2 equals 48 which is less than (<)97 not = to 97.
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Ginny is studying a population of frogs. She determines that the population is decreasing at an average rate of 3% per year. Whe
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Answer:

Step-by-step explanation:

This is exponential, of the form

y=a(b)^x where a is the initial amount and b is the growth/decay rate. If Ginny begins the study with 1200 frogs in the population, our a value is 1200.

If the population is decreasing 3% per year, it is decreasing 100% - 3% = 97% which, as a decimal, is .97. Our exponential function, then, is

y=1200(.97)^x

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3h - j when h = 8 and j = 11<br> plzzz help
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8 0
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
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