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3 years ago

What is the perimeter of a rectangle with a length of 2x + 4 and a width of 4x?

Mathematics

2 answers:

Orlov [11]3 years ago

8 0

Answer:

12x+8 or 4(3x+2)

Step-by-step explanation:

2x+4+4x+2x+4+4x

12x+8

4(3x+2)

Helen [10]3 years ago

7 0

I'll leave the picture to you; draw a rectangle and write 2x+4 along the length and 4x along the width. We don't really know the aspect ratio of the rectangle until we know x so any rectangle is fine.

The perimeter is the sum of all four side lengths,

p = (2x + 4) + 4x + (2x + 4) + 4x

Answer: p = 12x + 8

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Is anyone able to solve this?

Olenka [21]

Explanation:

To find x we need to find the unknown side that connects the two triangles using the Pythagorean theorem:

a² + b² = c² (c is always hypotenuse)

So:

a² + 6² = 9²

a² = 9² - 6²

a² = 81 - 36

a² = 45

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Now we do the same thing for the other triangle:

x² + 5² = sqrt45²

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8 months ago

What number should be added to both sides of the following equation to solve for x?

son4ous [18]

Answer:

7 is your answer

Step-by-step explanation:

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3 years ago

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MAVERICK [17]

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3 years ago

At noon, ship A is 130 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 20 km/h. How fast is

Hitman42 [59]

Answer:

answer = 12.87 km/h

Step-by-step explanation:

Given

Ship A is sailing east at 25 km/h = $\frac{dx}{dt}$

ship B is sailing north at 20 km/h = $\frac{dy}{dt}$

here x and y are the sailing at t = 4 : 00 pm for ship A and B respectively

so we get x = 4 ×25 =100 km/h

y = 4× 20 = 80 km/h

let z is the distance between the ships, we need to find $\frac{dz}{dt}$ at t = 4 hr

At noon, ship A is 130 km west of ship B (12:00 pm)

so equation will be

$z^2 = (130-x)^2 + y^2......................(i)\\put x = 100 and y = 80 \\\\we | | get \\$

$z^2 = 30^2 + 80^2\\z =\sqrt{7300} km/h$

derivative first equation w . r. to t we get

$2z\frac{dz}{dt} =$ $-2(130-x)\frac{dx}{dt}$ $+2y\frac{dy}{dt}$

$\frac{dz}{dt} =\frac{1}{z}[(x -130)\frac{dx}{dt} +y\frac{dy}{dt}]$

$\frac{dz}{dt} = \frac{( -20\times25 + 80\times20)}{\sqrt{7300} }$

$= \frac{1100}{85.44}\\ = 12.87km/h$

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2 years ago

What is 94.85 to the nearest tenth

allsm [11]

Answer:

94.90

Step-by-step explanation:

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2 years ago

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