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Verizon [17]
3 years ago
14

The length of a rectangle is twice its width and the area is equal to the area

Mathematics
2 answers:
Keith_Richards [23]3 years ago
6 0

Answer:

can one of yall help on mine

Step-by-step explanation:

Troyanec [42]3 years ago
6 0

The perimeter of the rectangle is 48 cm.

Step-by-step explanation:

Let the width of the rectangle = x

The length of the rectangle = 2x

so area =L X B = x X 2x =  2x^{2}

The each side of the square is 12 cm

so area = 12 X 12 sq cm = 144 sq cm

According to the problem,

2x^{2} = 144

or, x^{2} = 144/2 = 72

or, x = \sqrt{72}

= 8.4 = 8 (approx)

so length = 16 cm and width = 8 cm

Perimeter = 2 X ( L + B) = 2 X (16+8) = 48 cm

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C. The square root of 144
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Topic: The Quadratic Formula
Finger [1]

Answer:

Step-by-step explanation:

The quadratic formula for a equation of form

ax²+bx + c = 0 is

x= \frac{-b +- \sqrt{b^2-4ac} }{2a}

For the first equation,

x²+3x-4=0,

we can match that up with the form

ax²+bx + c = 0

to get that

ax² =  x²

divide both sides by x²

a=1

3x = bx

divide both sides by x

3 = b

-4 = c

. We can match this up because no constant multiplied by x could equal x² and no constant multiplied by another constant could equal x, so corresponding terms must match up.

Plugging our values into the equation, we get

x= \frac{-3 +- \sqrt{3^2-4(1)(-4)} }{2(1)} \\= \frac{-3+-\sqrt{25} }{2} \\ = \frac{-3+-5}{2} \\= -8/2 or 2/2\\=  -4 or 1

as our possible solutions

Plugging our values back into the equation, x²+3x-4=0, we see that both f(-4) and f(1) are equal to 0. Therefore, this has 2 real solutions.

Next, we have

x²+3x+4=0

Matching coefficients up, we can see that a = 1, b=3, and c=4. The quadratic equation is thus

x= \frac{-3 +- \sqrt{3^2-4(1)(4)} }{2(1)}\\= \frac{-3 +- \sqrt{9-16} }{2}\\= \frac{-3 +- \sqrt{-7} }{2}\\

Because √-7 is not a real number, this has no real solutions. However,

(-3 + √-7)/2 and (-3 - √-7)/2 are both possible complex solutions, so this has two complex solutions

Finally, for

4x² + 1= 4x,

we can start by subtracting 4x from both sides to maintain the desired form, resulting in

4x²-4x+1=0

Then, a=4, b=-4, and c=1, making our equation

x=\frac{-(-4) +- \sqrt{(-4)^2-4(4)(1)} }{2(4)} \\= \frac{4+-\sqrt{16-16} }{8} \\= \frac{4+-0}{8} \\= 1/2

Plugging 1/2 into 4x²+1=4x, this works as the only solution. This equation has one real solution

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