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egoroff_w [7]
3 years ago
9

Please solve for me it's urgent. A rectangle has 10 and 6cm each. find the area of a square with the same perimeter as that of t

he rectangle
​
Mathematics
1 answer:
elena55 [62]3 years ago
5 0

Answer:

64 cm²

Step-by-step explanation:

Firstly,let us lay down the clues;

So, we have been given the Length and Width of the rectangle.

Rectangle;

<em>Length</em><em>=</em><em> </em><em>10</em><em> </em><em>cm</em>

<em>Width</em><em>=</em><em>6</em><em> </em><em>cm</em>

We know that

Perimeter= 2L + 2W

Perimeter= 10+10+6+6

Perimeter =32 cm

In the questionnaire, it is mentioned that a square has the same perimeter as the rectangle. That means, the perimeter of the square is similar to the one of the rectangle which is 32 cm.

So,

Perimeter of square =32 cm

Length of one side of the square =32÷4

=8 cm

Area= L × W

Area= 8 cm× 8 cm

Area= 64 cm²

Hope it helps.

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What is the slope of a line that passes through the points (4, - 3) and (- 2, 9)
oee [108]

Answer:

-2

Step-by-step explanation:

slope = y2 - y1  / x2 - x1

= 9 - (-3)  /   -2 -4

= 12 / -6

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3 0
2 years ago
There are three local factories that produce radios. Each radio produced at factory A is defective withprobability .02, each one
diamong [38]

Answer:

The probability is 0.02667

Step-by-step explanation:

Let's call D1 the event that the first radio is defective and D2 the event that the second radio is defective.

So, if we select both radios any factory, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

P(D2/D1) = P(D2∩D1)/P(D1)

Taking into account that 0.02 is the probability that a radio produced at factory A is defective, P(D2/D1) for factory A is:

P(D2/D1)_A=\frac{0.02*0.02}{0.02} =0.02

At the same way, if both radios are from factory B, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

P(D2/D1)_B=\frac{0.01*0.01}{0.01} =0.01

Finally, if both radios are from factory C, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

P(D2/D1)_C=\frac{0.05*0.05}{0.05} =0.05

So, if the radios are equally likely to have been any factory, the probability to select both radios from any of the factories A, B or C are respectively:

P(A)=1/3

P(B)=1/3

P(C)=1/3

Then, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

P(D2/D1)=P(A)P(D2/D1)_A+P(B)P(D2/D1)_B+P(C)P(D2/D1)_C

P(D2/D1) = (1/3)*(0.02) + (1/3)*(0.01) + (1/3)*(0.05)

P(D2/D1) = 0.02667

6 0
3 years ago
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alekssr [168]

Answer:

x = 0 and y = 4

x = 4 and y = 0

thus: 2 solutions

Step-by-step explanation:

8 0
3 years ago
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