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3 years ago

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Mathematics

1 answer:

MAXImum [283]3 years ago

6 0

Answer: A. Figure ABCD is similar to figure A'B'C'D'.

Step-by-step explanation:

Given : Figure ABCD is reflected about the x-axis to obtain figure A'B'C'D'.

Since a reflection is a rigid transformation which produces congruent images.

⇒ Figure ABCD must be congruent to figure A'B'C'D'.

⇒ Figure ABCD has same size as figure A'B'C'D'.

⇒ Figure ABCD must be similar to figure A'B'C'D'. [ ∵ If two figures are congruent then they are also similar.]

Reflection preserves side length and measure of angles .

⇒ Corresponding angles and side lengths of figure ABCD must be congruent to figure A'B'C'D'.

⇒ The measure of angle D is equal to the measure of angle D'.

Hence, the statement best describes the relationship between the two figures is :-

A. Figure ABCD is similar to figure A'B'C'D'.

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3 years ago

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3 years ago

Given (21,7) and (x,<br> 1), find all x such that the distance between these two points is 17.

Dmitriy789 [7]

Answer:

21+/-sqrt(253)=x

So one value for x is 21+sqrt(253)

and another is 21-sqrt(253)

Problem:

Given (21,7) and (x,1), find all x such that the distance between these two points is 17.

Step-by-step explanation:

Change in x is x-21

Change in y is 7-1=6

distance^2=(change in x)^2+(change in y)^2

17^2=(x-21)^2+(6)^2

289=(x-21)^2+36

Subtract 36 on both sides:

289-36=(x-21)^2

253=(x-21)^2

Take square root of both sides:

+/-sqrt(253)=x-21

Add 21 on both sides:

21+/-sqrt(253)=x

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3 years ago

Hey if you're willing to help me with this, it will be very much appreciated, thanks maties and mates.

Nastasia [14]

Answer:

Width=37in

Length=55in

Step-by-step explanation:

Let x in be the width dimension, the length dimension will be x+18 in. Area of a rectangle is calculated as A=lw. Given the area as2143 sq in, x can be calculated as:

$Area= x(18+x)\\\\=18x+x^2\\\\$

#Substitute for every x in our function to find the true value of x:

$x=37\\\\A=x^2+18x\\\\=37^2+18\times37\\\\=2035 \ in^2$ #l=37+18=55in

for x=38:

$x=38\\\\A=x^2+18x\\\\=38^2+18\times38\\\\=2128 \ in^2$ #l=38+18=56in

For x=39:

$x=39\\\\A=x^2+18x\\\\=39^2+18\times39\\\\=2223 \ in^2$ #l=39+1854in

From the calculations above, A television with a width of 37 in and length 55in has its area closest to 2143 sq in.

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3 years ago