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CaHeK987 [17]
3 years ago
14

What is the volume of a box that is 5 feet long, 6 feet wide and 10 feet tall

Mathematics
1 answer:
WITCHER [35]3 years ago
7 0

Answer:

The volume is 300 square feet.

Step-by-step explanation:

Volume is calculated by multiplying as such:

l \times w \times h

so, to calculate this, you would insert the parameters of the equation like this:

5 \times 6 \times 10

doing it step by step, it is:

5 \times 6 = 30

30 \times 10 = 300

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What is the missing length?
TEA [102]
Answer:  13.722 km ;  or, write as:  13  13/18 km .
___________________________________________________
Explanation:
___________________________________________________
Area = Length * width ; 
___________________________________________________
or, write as:  A = L * w ;
________________________

Given: A = 247 km² ; 
           
          L = 18 km ; 
 
          w = "y" ; 
________________________
 Find:  "y"
______________________________
  A = L * w ; 
______________________________
Plug in our values:
______________________________
  247 km² = 18 km * "y" ;  solve for "y" (in units of "km") ;

_____________________________
        18 y = 247 ; 
___________________________________________________________
Divide each side of the equation by "18"; to isolate "y" on one side of the equation; and to solve for "y" :
___________________________________________________________
        18 y / 18 = 247 / 18 ;
_____________________________________________
  to get:
_____________________________________________
              y = 13.7222222222222222...... km ;  round to: 13.722 km

                   or;  y = 13  13/18 km .
_______________________________________________
3 0
3 years ago
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nignag [31]

Answer:

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3 0
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Write an equation to find the number. Solve. Twenty-four less than 4 times a number is equal to 30 more than -2 times the number
Amanda [17]
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a

Step-by-step explanation:

6 0
3 years ago
(with steps please) Find the inverse Laplace transform, f(t), of the function: s/((s^2+9)*(s^2+16))
sveta [45]

Answer:

Step-by-step explanation:

Given is a Laplace transform of f(t) as

F(s) = \frac{s}{(s^2+9)(s^2+16)}

Let

\frac{s}{(s^2+9)(s^2+16)}=\frac{As+B}{(s^2+9)}+\frac{Cs+D}{(s^2+16)}

Solving we get \:s=s^3\left(A+C\right)+s^2\left(B+D\right)+s\left(16A+9C\right)+\left(16B+9D\right)

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Solving B=0=D

A=\frac{1}{7} \\C=\frac{-1}{7}

Hence we have

f(t) = inverse Laplace of

\frac{1}{7}[ \frac{s}{(s^2+9)}-\frac{s}{(s^2+16)}]

=\frac{1}{7} (cos 3t-cos 4t)

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