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il63 [147K]
3 years ago
12

What are the zeros of this graph

Mathematics
1 answer:
kotegsom [21]3 years ago
3 0

The zeroes ( where the graph cuts the x axis) ar (-2,0) AND (2,0)

tHE FACTOrIAL FORM IS (x - 2)(x + 2)

Its B

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I NEED HELP CROWN FOR RIGHT ANSWER
natta225 [31]

Answer:

D. $81,728

Step-by-step explanation:

Let's first convert pounds to ounces.

4 pounds x (16 ounces / pound) = <u>64 ounces.</u> (Remember that when we're converting from a larger unit, pounds, to a smaller one, ounces, we'll have more ounces than pounds, so this answer makes sense!)

Let's now get the cost of the gold.

64 oz x ($1277 / oz) = <u>$81,728.</u> (Wow, that's a lot of gold!)

5 0
2 years ago
NEED DONE ASAP!!
almond37 [142]

Answer:

A. 3x^2 + 4 = 52

     3x^2 = 52 - 4

     3x^2 = 48

       x^2 = 48/3

       x^2 = 16

        x    = \sqrt{16}

        x = 4

B. x^2 - 121 = 0

     x^2  = 121

       x = \sqrt{121}

       x = 11

C. (b + 4 )2 = 49

    2b + 8 = 49

    2b  = 49 - 8

     2b = 41

        b = 41/2

3 0
2 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}

                       

                                                 

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