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adell [148]
3 years ago
6

A square piece of cardboard with each side 30 inches long hasa

Mathematics
1 answer:
Murljashka [212]3 years ago
7 0

Answer: 5 inches

Step-by-step explanation:

V = S * (30-2S)^2

V = S(900-120s+4s^2)

Volume = 4s^3 -120s^2 +900s

Find local maximum from graph at 5 (or take derivative = 0)

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-x + 23 < 2 - 2(x – 8)​
mihalych1998 [28]

Answer:

x<-5

Step-by-step explanation:

that's the answer

6 0
3 years ago
How do you graph a circle x^2 + y^2=25 and the line is given y=2
Kazeer [188]
\bf \textit{equation of a circle}\\\\ &#10;(x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2&#10;\qquad &#10;\begin{array}{lllll}&#10;center\ (&{{ h}},&{{ k}})\qquad &#10;radius=&{{ r}}&#10;\end{array}\\\\&#10;-------------------------------\\\\&#10;x^2+y^2=25\implies (x-0)^2+(y-0)^2=5^2

so notice above, the circle is centered at h,k or 0,0, the origin, and has a ratio of 5.

and y = 2, is just a horizontal line, check the picture below.

7 0
3 years ago
13 + (4 + 10) = 27 the numbers and the parentheses to create an equivalent addition sentence that illustrates the associative pr
rewona [7]
This isn't a question....
4 0
3 years ago
Which is the simplified form of the expression (6^-2•6^5)^-3
Arisa [49]

Answer:

1

Step-by-step explanation:

( {6}^{ - 2}  \times  {6}^{5} ) ^{ - 3}  \\  = ( {6}^{3} )^{ - 3}  \\  =  {6}^{0}  \\  = 1

6 0
3 years ago
Read 2 more answers
PLEASE HELP ASAP!!! I don’t know how to do this problem or where to start! How do I solve this?
Leto [7]

Answer:

W = 4.95

Step-by-step explanation:

You want to start by writing down what you know, and forming a system of equations.

L= length W= width

2L+2W=14.7    

L= 2.4

On the left side of the equation, you're adding all your side lengths, and on the right, is the total perimeter. (Also could be written L+L+W+W = 14.7)

You would then substitute L from the bottom equation into the top equation to get:

2(2.4) +2W=14.7

Solving:

4.8+2w=14.7

W= 4.95

To check your answer simply add all the sides together and make sure it equals your perimeter. You can also plug W and L back into the original equation.

7 0
3 years ago
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