The perimeter of a rectangle is 30 cm. The width of the rectangle is 7 cmWhat is the area of the rectangle?

A) 56 cm

B) 49 cm

C) 14 cm

Answers

Answer 1

Answer: Let's say the length of the rectangle is L.

We have P= (L+W)*2
⇒ 30 cm= (L+ 7 cm)*2
⇒ L+ 7 cm= 15 cm
⇒ L= 8 cm.

S= L*W= 8 cm* 7 cm= 56 cm^2.

The answer is A) 56 cm^2.

Hope this helps~

Answer 2

Answer: perimeter = 2(l+w)
30 = 2(l+7)
30 = 2l + 14
30 - 14 = 2l
16 = 2l
8 = l

l = 8
w = 7

area = l x w = 8 x 7 = 56 cm²

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Given that / and /, which postulate or theorem can be used to show that all 3 center triangles in the Antigua and Barbuda flag are similar?

Answers

Given that we assume that all the bases of the triangles are parallel.

We can use AAA or Angle-Angle-Angle to prove that these triangles are similar.

Each parallel line creates the same angle when intersecting with the same side.
For example:
The bases of each triangle cross the left side of all the triangle.
Each angle made by the intersecting of the the parallel base and the side are the same.
Thus, each corresponding angle of all the triangles are congruent.
If these angles are congruent, then we have similar triangles.

Actually, the answer is ASA postulate!!!! I just took the county geometry quiz and got it correct

Help!! Can someone put me out my misery

Answers

Answer: 0.14

Step-by-step explanation:

1) Find the total #of seniors. To do this add up 25+5+5 = 35.

2)Since we know the number of seniors that work just do 5/35. This simplifies to 0.14 rounded to the nearest hundreth.

Factorise:a4+2a3b-2ab3-b4..

Answers

$a^4+2a^3b-2ab^3-b^4\n\n=a^4-b^4+2a^3b-2ab^3\n\n=(a^2-b^2)(a^2+b^2)+2ab(a^2-b^2)\n\n=(a^2-b^2)(a^2+b^2+2ab)\n\n=(a-b)(a+b)(a+b)^2\n\n=(a-b)(a+b)(a+b)(a+b)$

$(x+y)^2=x^2+2xy+y^2\n\nx^2-y^2=(x-y)(x+y)$

What’s the answer???