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Aleonysh [2.5K]
3 years ago
5

Patricia purchased x meters of fencing. She originally intended to use all of the fencing to enclose a square region, but later

decided to use all of the fencing to enclose a rectangular region with length y meters greater than its width. In square meters, what is the positive difference between the area of the square region and the area of the rectangular region?
Mathematics
1 answer:
sergejj [24]3 years ago
5 0

Answer:

  (y^2)/4 square meters

Step-by-step explanation:

For a perimeter length of x, the side of a square will be x/4 and its area will be (x/4)^2.

If one side of the square is shortened by y/2 and the adjacent side is lengthened by y/2, then the difference in side lengths will be y. The area of the resulting rectangle will be ...

  (x/4 -y/2)(x/4 +y/2) = (x/4)^2 -(y/2)^2

That is, the difference in area between the square and the rectangle is ...

  (x/4)^2 - ((x/4)^2 -(y/2)^2) = (y/2)^2 = y^2/4

The positive difference between the area of the square region and the area of the rectangular region is y^2/4 square meters.

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3 years ago
two dice are rolled until a sum of either 5 or 7 occurs. estimate the probability that a 5 occurs first g
Ierofanga [76]

Answer:

The correct answer is "\frac{2}{5}". A further solution is provided below.

Step-by-step explanation:

According to the question,

The probability of getting a 5 will be:

= \frac{4}{36}

= \frac{1}{9}

The probability of getting a 7 will be:

= \frac{6}{36}

= \frac{1}{6}

The probability of not a 5 nor a 7 will be:

= 1-[\frac{4}{36} +\frac{4}{36} ]

= \frac{26}{36}

= \frac{13}{18}

Now,

For roll of 2 dice,

⇒ P=\frac{1}{9} +{\frac{1}{9}\times (\frac{13}{18} ) }+{(\frac{1}{9} )\times (\frac{13}{18} )^2}+{\frac{1}{9}\times (\frac{13}{18} )^3 }+...

        =\frac{1}{9} [1+\frac{13}{18} +\frac{13^2}{18^2} +\frac{13^3}{18^3} +...]

        =\frac{1}{9} [\frac{1}{1-\frac{13}{18} } ]

        =\frac{1}{9} [{\frac{1}{\frac{5}{8} } ]

        =\frac{2}{5}

6 0
3 years ago
What are the foci of the hyperbola with equation 5y^2-4x^2=20?
shusha [124]
The foci of the hyperbola with equation 5y^2-4x^2=20 will be given as follows:
divide each term by 20
(5y^2)/20-(4x^2)/20=20/20
simplifying gives us:
y^2/4-x^2/5=1
This follows the standard form of the hyperbola
(y-k)²/a²-(x-h)²/b²=1
thus
a=2, b=√5 , k=0, h=0
Next we find c, the distance from the center to a focus.
√(a²+b²)
=√(2²+(√5)²)
=√(4+5)
=√9
=3
the focus of the hyperbola is found using formula:
(h.h+k)
substituting our values we get:
(0,3)
The second focus of the hyperbola can be found by subtracting c from k
(h,k-c)
substituting our values we obtain:
(0,-3)

Thus we have two foci
(0,3) and (0,-3)
5 0
3 years ago
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Answer: Well if you divide 4200 by 42, you will get 100, which is a multiple of 10

Step-by-step explanation:

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