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Kisachek [45]
2 years ago
7

Which of the following sets of numbers could not represent the three sides of a triangle?

Mathematics
1 answer:
leonid [27]2 years ago
5 0

Answer:

Correct answer is option C

( 15 , 21 , 33 )

Step-by-step explanation:

hope it helps

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The dimensions of a rectangle are such that its length is 3 in. more than its width. If the length were doubled and if the width
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Answer:

The length of the rectangle is 11 inch and its width is 8 inch.

Step-by-step explanation:

Given:

Rectangle (R_1) and rectangle (R_2).

Let the width of the rectangle be 'x' inch.

Its length = '(x+3)' inch

Area of the rectangle = (A_1)

⇒A_1 = (x)(x+3)

Now for rectangle (R_2) its width is decreased by 1 inch and length is doubled,the area (A_2) will increased by 66 sq-inch.

So,

A_2=A_1+66

According to the question.

⇒ Area of rectangle (R_2)  with width '(x-1)' and length '2(x+3)'='(2x+6)' .

⇒ A_2=(x-1)(2x+6)

Plugging the values of A_2 in terms of A_1.

⇒ A_1+66= (x-1)(2x+6)

Plugging  A_1 = (x)(x+3) into the equation.

⇒ (x)(x+3)+66=(x-1)(2x+6)

⇒ x^2+3x+66=2x^2+6x-2x-6

⇒ x^2-2x^2+3x-6x+2x+6+66=0

⇒ -x^2-x+72=0

Now using quadratic formula we can find the value of 'x'.

Quadratic formula, (x)= \frac{\pm b+\sqrt{b^2-4ac} }{2a}

x= -9 and x= 8

Discarding the negative value.

Our width  = 'x' = 8 inches.

So the length of the rectangle = 11 inches and width of the rectangle = 8 inches.

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Answer:

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