Correct question :
If the perimeters of each shape are equal, which equation can be used to find the value of x? A triangle with base x + 2, height x, and side length x + 4. A rectangle with length of x + 3 and width of one-half x. (x + 4) + x + (x + 2) = one-half x + (x + 3) (x + 2) + x + (x + 4) = 2 (one-half x) + 2 (x + 3) 2 (x) + 2 (x + 2) = 2 (one-half x) + 2 (x + 3) x + (x + 2) + (x + 4) = 2 (x + 3 and one-half)
Answer: (x + 2) + x + (x + 4) = 2 (one-half x) + 2 (x + 3)
Step-by-step explanation:
Given the following :
A triangle with base x + 2, height x, and side length x + 4 - - - -
b = x + 2 ; a = x ; c = x + 4
Perimeter (P) of a triangle :
P = a + b + c
P =( x + 2) + x + (x + 4) - - - (1)
A rectangle with length of x + 3 and width of one-half x
l = x + 3 ; w = 1/2 x
Perimeter of a rectangle (P) = 2(l+w)
P = 2(x+3) + 2(1/2x)
If perimeter of each same are the same ; then;
(1) = (2)
(x + 2) + x + (x + 4) = 2(x+3) + 2(1/2x)
Answer:
A, C, E
Step-by-step explanation:
Only letters T, A, and X on the triangle can be used to describe it, regardless of the order they are in.
The <em>proposed</em> design of the atrium (<em>V < V'</em>) is possible since its volume is less than the <em>maximum possible</em> atrium.
<h3>Can this atrium be built in the rectangular plot of land?</h3>
The atrium with the <em>maximum allowable</em> radius (<em>R</em>), in feet, is represented in the image attached. The <em>real</em> atrium is possible if and only if the <em>real</em> radius (<em>r</em>) is less than the maximum allowable radius and therefore, the <em>real</em> volume (<em>V</em>), in cubic feet, must be less than than <em>maximum possible</em> volume (<em>V'</em>), in cubic feet.
First, we calculate the volume occupied by the maximum allowable radius:
<em>V' =</em> (8 · π / 3) · (45 ft)³
<em>V' ≈</em> 763407.015 ft³
The <em>proposed</em> design of the atrium (<em>V < V'</em>) is possible since its volume is less than the <em>maximum possible</em> atrium.
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Answer:
396cm I think.
Step-by-step explanation:
10 x (27 - 8) + 9 x 24