The line VS is middle line of that trapezoid, so you know that it is
VS= (WR+UT)/2
so we will get that it is:
<span>40= (x^2+2 +2x^2+3)/2
x = 5
Therefore,
x^2 + 2
5^2 + 2
25 +2
27 <----- OPTION 2
Hope this answers the question.</span>
Answer:
(7 X 3) x 6 = 7 x (3 X 6)
Step-by-step explanation:
Answer:
Part 1) m∠EFG=94°
Part 2) m∠GFH=86°
Step-by-step explanation:
we know that
m∠EFG+m∠GFH=180° -----> by linear pair (given problem)
we have
m∠EFG=3n+22
m∠GFH=2n+38
substitute the values
(3n+22)°+(2n+38)°=180°
Solve for n
(5n+60)=180
5n=180-60
5n=120
n=24
<em>Find the measure of angle EFG</em>
m∠EFG=3n+22
substitute the value of n
m∠EFG=3(24)+22=94°
<em>Find the measure of angle GFH</em>
m∠GFH=2n+38
substitute the value of n
m∠GFH=2(24)+38=86°
The correct answer is: [B]: "40 yd² " .
_____________________________________________________
First, find the area of the triangle:
The formula of the area of a triangle, "A":
A = (1/2) * b * h ;
in which: " A = area (in units 'squared') ; in our case, " yd² " ;
" b = base length" = 6 yd.
" h = perpendicular height" = "(4 yd + 4 yd)" = 8 yd.
___________________________________________________
→ A = (1/2) * b * h = (1/2) * (6 yd) * (8 yd) = (1/2) * (6) * (8) * (yd²) ;
= " 24 yd² " .
___________________________________________________
Now, find the area, "A", of the square:
The formula for the area, "A" of a square:
A = s² ;
in which: "A = area (in "units squared") ; in our case, " yd² " ;
"s = side length (since a 'square' has all FOUR (4) equal side lengths);
A = s² = (4 yd)² = 4² * yd² = "16 yd² "
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Now, we add the areas of BOTH the triangle AND the square:
_________________________________________________
→ " 24 yd² + 16 yd² " ;
to get: " 40 yd² " ; which is: Answer choice: [B]: " 40 yd² " .
_________________________________________________
Answer:
Hello,
Step-by-step explanation:
![A=(1,2)\\B=(0,-1)\\\overrightarrow{AB}=((0,-1)-(1,2)=(-1,-3)\ ||\overrightarrow{AB}||^2=1+9=10\\\overrightarrow{BC}=((3,-2)-(0,-1)=(3,-1)\ ||\overrightarrow{BC}||^2=9+1=10\\\\Triangle\ is\ isosceles.\\\\\overrightarrow{AB}.\overrightarrow{BC}=(-1,-3)*\left[\begin{array}{c}3\\-1\end{array}\right] =-3+3=0\\\\Triangle \ is\ right.\\\\](https://tex.z-dn.net/?f=A%3D%281%2C2%29%5C%5CB%3D%280%2C-1%29%5C%5C%5Coverrightarrow%7BAB%7D%3D%28%280%2C-1%29-%281%2C2%29%3D%28-1%2C-3%29%5C%20%7C%7C%5Coverrightarrow%7BAB%7D%7C%7C%5E2%3D1%2B9%3D10%5C%5C%5Coverrightarrow%7BBC%7D%3D%28%283%2C-2%29-%280%2C-1%29%3D%283%2C-1%29%5C%20%7C%7C%5Coverrightarrow%7BBC%7D%7C%7C%5E2%3D9%2B1%3D10%5C%5C%5C%5CTriangle%5C%20is%5C%20isosceles.%5C%5C%5C%5C%5Coverrightarrow%7BAB%7D.%5Coverrightarrow%7BBC%7D%3D%28-1%2C-3%29%2A%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D3%5C%5C-1%5Cend%7Barray%7D%5Cright%5D%20%3D-3%2B3%3D0%5C%5C%5C%5CTriangle%20%5C%20is%5C%20right.%5C%5C%5C%5C)
