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Marysya12 [62]
3 years ago
12

The length of the sides of a square measure 2x-5. The length of a rectangle measures 2x, and the width measures x + 2. For what

value of x is the perimeter of the square the same as the perimeter of the rectangle?
Mathematics
1 answer:
vladimir1956 [14]3 years ago
3 0

Answer:

x = 12

Step-by-step explanation:

Perimeter of a square:

The perimeter of a square of side x is given by:

P_s = 4x

Perimeter of a rectangle:

The perimeter of a rectangle of length l and width w is given by:

P_r = 2(l + w)

The length of the sides of a square measure 2x-5.

This means that the perimeter of the square is:

P_s = 4(2x - 5) = 8x - 20

The length of a rectangle measures 2x, and the width measures x + 2.

This means that the perimeter of the rectangle is:

P_r = 2(2x + x + 2) = 2(3x + 2) = 6x + 4

For what value of x is the perimeter of the square the same as the perimeter of the rectangle?

This is x for which:

P_s = P_r

So

8x - 20 = 6x + 4

8x - 6x = 20 + 4

2x = 24

x = \frac{24}{2}

x = 12

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(4x2 + 6) + (2x2 + 6x + 3)
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3 years ago
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nata0808 [166]

Answer:

1125 m

Step-by-step explanation:

Given equation:

h=-5t^2+150t

where:

  • h = height (in metres)
  • t = time (in seconds)

<u>Method 1</u>

Rewrite the equation in vertex form by completing the square:

h=-5t^2+150t

\implies h=-5t^2+150t-1125+1125

\implies h=-5(t^2-30t+225)+1125

\implies h=-5(t-15)^2+1125

The vertex (15, 1125) is the turning point of the parabola (minimum or maximum point).  As the leading coefficient of the given equation is negative, the parabola opens downward, and so vertex is the maximum point.  Therefore, the maximum height is the y-value of the vertex: 1125 metres.

<u>Method 2</u>

Differentiate the function:

\implies \dfrac{dh}{dt}=-10t+150

Set it to zero and solve for t:

\implies -10t+150=0

\implies 10t=150

\implies t=15

Input found value of t into the original function and solve for h:

\implies -5(15)^2+150(15)=1125

Therefore, the maximum height is 1125 metres.

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