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stiv31 [10]
3 years ago
15

Ben has two boxes. The red box has a volume of 46 cubic feet. The blue box has a height of 3 feet, a length of 7 feet, and a wid

th of 4 feet. How much greater is the blue box than the red box?
Mathematics
1 answer:
Juliette [100K]3 years ago
8 0

Answer:

The blue box is 38 cubic feet grater than the red box.

Step-by-step explanation:

Cuboid:

  • It is a three dimensional shape.
  • It has 6 faces.
  • The surface area of a cuboid is = 2(lb+bh+lh), where l = length, b= width, h= height.
  • The volume of a cuboid is = l×b×h

Given that,

Ben has two boxes.

The volume of the red box is 46 cubic feet.

The dimensions of the blue box is 7 feet by 4 feet by 3 feet.

The volume of the blue box is = length ×width × height

                                                  =(7×4×3) cubic feet

                                                  = 84 cubic feet

The blue box is (84-46) cubic feet =38 cubic feet grater than the red box.

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Answer:

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Step-by-step explanation:

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Answer:

$460.20

Step-by-step explanation:

499*15% = 74.85 (total discount price)

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nata0808 [166]

Answer:

Hi there!

Your answer is:

B. 1.5

Step-by-step explanation:

The square root of 2.25 = 1.5

Check your work!

1.5×1.5 = 2.25!!✓

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7 0
3 years ago
Find the solution to this system of equations <br> 3x+2y+3z=3 4x-5y+7z=1 2x+3y-2z=6 <br> x=? Y=? Z=?
Ede4ka [16]

Answer:

The solution is x=2,\ y=0,\ z=-1.

Step-by-step explanation:

You are given the system of three equations:

\left\{\begin{array}{l}3x+2y+3z=3\\4x-5y+7z=1\\2x+3y-2z=6\end{array}\right.

Multiply the first equation by 4, the second equation by 3 and subtract them. Then multiply the third equation by 2 and subtract it from the second equation:

\left\{\begin{array}{l}3x+2y+3z=3\\4(3x+2y+3z)-3(4x-5y+7z)=4\cdot 3-3\cdot 1\\4x-5y+7z-2(2x+3y-2z)=1-2\cdot 6\end{array}\right.\Rightarrow \\\\\left\{\begin{array}{l}3x+2y+3z=3\\12x+8y+12z-12x+15y-21z=12-3\\4x-5y+7z-4x-6y+4z=1-12\end{array}\right.

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\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\-11y+11z=-11\end{array}\right.\Rightarrow \left\{\begin{array}{l}3x+2y+3z=3\\23y-9z=9\\y-z=1\end{array}\right.

Multiply the third equation by 23 and subtract it from the second equation:

\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\23y-9z-23(y-z)=9-23\cdot 1\end{array}\right.\Rightarrow \left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\23y-9z-23y+23z=9-23 \end{array}\right.

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The solution is x=2,\ y=0,\ z=-1.

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The value of b is -21.
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