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3 years ago

How would you solve -92+x=151-8x

Mathematics

2 answers:

frosja888 [35]3 years ago

6 0

-92+x=151-8x
x=-92+151-8x
x+8x=59
9x=59
x=6.5
I don't know if this is completely correct

lyudmila [28]3 years ago

5 0

First group like terms together to get
x + 8x = 151 +92

then add.
9x = 243

divide both sides by 9
9x/9 = 243/9

x = 27

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Rectangle 1 has length x and width y. Rectangle 2 is made by multiplying each dimension of Rectangle 1 by a factor of K, where k

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$\bold{\huge{\underline{\pink{ Solution }}}}$

<h3>Given :- </h3>

Rectangle 1 has length x and width y
Rectangle 2 is made by multiplying each dimensions of rectangle 1 by a factor of k
Where, k > 0

<h3>Answer 1 :-</h3>

Yes, The rectangle 1 and rectangle 2 are similar .

<h3>According to the similarity theorem :-</h3>

If the ratio of length and breath of both the triangles are same then the given triangles are similar.

Let's Understand the above theorem :-

The dimensions of rectangle 1 are x and y

Now, 

Rectangle 2 is made by multiplying each dimensions of rectangle 1 by a factor of k .

Let assume the value of K be 5

Therefore, 

The dimensions of rectangle 2 are

$\sf{ 5x \:and \:5y }$

Now, The ratios of dimensions of both the rectangle :-

$\bold{Rectangle 1 = Rectangle 2}$

$\bold{\dfrac{ x }{y}}{\bold{ = }}{\bold{\dfrac{5x}{5y}}}$

$\bold{\blue{\dfrac{ x }{y}}}{\bold{\blue{ = }}}{\bold{\blue{\dfrac{x}{y}}}}$

From above, 

We can conclude that the ratios of both the rectangles are same

Hence , Both the rectangles are similar

<h3>Answer 2 :- </h3>

Here, 

We have to proof that, the

Perimeter of rectangle 2 = k(perimeter of rectangle 1 )

In the previous questions, we have assume the value of k = 5

<h3>Therefore, </h3>

According to the question,

Perimeter of rectangle 1

$\sf{ = 2( length + Breath) }$

$\bold{\pink{= 2( x + y ) }}$

Thus, The perimeter of rectangle 1

Perimeter of rectangle 2

$\sf{ = 2( length + Breath) }$

$\sf{ = 2(5x + 5y) }$

$\sf{ = 2 × 5( x + y) }$

$\bold{\pink{= 10(x + y) }}$

Thus, The perimeter of rectangle 2

According to the given condition :-

Perimeter of rectangle 2 = k( perimeter of rectangle 1 )

Subsitute the required values,

$\sf{ 2(x + y) = 10(x + y)}$

$\bold{\pink{2x + 2y = 5(2x + 2y) }}$

From Above, 

We can conclude that the, Perimeter of rectangle 2 is 5 times of perimeter of rectangle 1 and we assume the value of k = 5.

Hence, The perimeter of rectangle 2 is k times of rectangle 1

<h3>Answer 3 :</h3>

Here, 

We have to proof that ,

The area of rectangle 2 is k² times of the area of rectangle 1.

That is, 

Area of rectangle 1 = k²( Area of rectangle)

<h3>Therefore, </h3>

According to the question,

Area of rectangle 1

$\sf{ = Length × Breath }$

$\sf{ = x × y }$

$\bold{\red{= xy }}$

Area of rectangle 2

$\sf{ = Length × Breath }$

$\sf{ = 5x × 5y }$

$\bold{\red{ = 25xy }}$

According to the given condition :-

Area of rectangle 1 = k²( Area of rectangle)

$\sf{ xy = 25xy }$

$\bold{\red{xy = (5)²xy }}$

From Above, 

We can conclude that the, Area of rectangle 2 is (5)² times of area of rectangle 1 and we have assumed the value of k = 5

Hence, The Area of rectangle 2 is k times of rectangle 1 .

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