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3 years ago

find the missing length x in these keloids cuboids given their surface area. Plz tell me how you did it too. i realy need help.

Mathematics

1 answer:

emmainna [20.7K]3 years ago

8 0

Surface area =2*(3*x)+2*(1*x)+2*(3*1)=6x+2x+6=8x+6
70=8x+6
64=8x
x=8

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3 years ago

The length of one side of a triangular lot is 6 m less than 3 times the length of the second side. The third side is 8 m longer

Tanya [424]

second side = x

first side = 3x - 6

third side = (3x - 6) + 8

The perimeter = 80m and is equal x + (3x - 6) + [(3x - 6) + 8]

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3 years ago

A 4-lb. force acting in the direction of (vector) 4,-2 moves an object just over 7ft from point (0,4) to (5,-1). Find the work d

Tcecarenko [31]

To solve this problem, we have to find the net displacement and the net force and the multiply the dot product together and get the work done.

The work done on moving the object is 27ft*lbs

<h3>Work done in moving the object from point A to point B</h3>

To find the work done on this object, let's find the net force on the object.

Data;

force = 4lb
direction = 4, -2
displacement = 7ft
direction = (0, 4) to (5,1)

The unit vector of the force is

$\sqrt{4^2 +(-2)^2} =\sqrt{16 + 4} = \sqrt{20}$

$\frac{4}{\sqrt{20} }, \frac{-2}{\sqrt{20} }$

The net force acting on the object is

$F = 4(\frac{4}{\sqrt{20} }, \frac{-2}{\sqrt{20} })\\F= (\frac{16}{\sqrt{20} }, \frac{-8}{\sqrt{20} } )$

The displacement on the object is 7ft through (0,4) to (5, -1)

The unit vector on displacement is

$\sqrt{5^2 + (-1-4)^2} = \sqrt{25+25} = \sqrt{50}$

$\frac{5}{\sqrt{50} }, \frac{-5}{\sqrt{50} }$

The net displacement will be

$7(\frac{5}{\sqrt{50} }, \frac{-5}{\sqrt{50} }) = \frac{35}{\sqrt{50} }, \frac{-35}{50}$

The work done will be F.d

$w = f. d \\$

$w = (\frac{16}{\sqrt{20} }, \frac{-8}{\sqrt{20} } ) * \frac{35}{\sqrt{50} }, \frac{-35}{50}\\w = 17.71+ 8.854\\w = 26.567 = 27ft*lbs$

The work done on moving the object is 27ft*lbs

Learn more on work done on an object here;

brainly.com/question/26152883

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