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 "
_________________________________________________
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 " .
_________________________________________________
PART1:
First Combination:
Pizza ($7) + Chicken Strips ($6) + Biscuits ($3) + Grapes ($4) = $20
Second Combination:
Dog Food ($13) + Bread ($3) + Crackers ($2) + Broccoli ($2) = $20
Third Combination:
Shampoo ($4) + Tissues ($3) + Pizza ($7) + Eggs ($3) + Biscuits ($3) = $20
PART 2:
First Combination:
$7.20 + $5.70 + $2.90 + $3.70 = $19.60
No, I wouldn’t have gone over the limit
Second Combination:
$13.40 + $3.50 + $2.00 + $1.90 = $20.80
Yes, I would have gone over the limit
Third Combination:
$3.50 + $2.60 + $7.20 + $2.50 + $2.90 = $18.70
No, I wouldn’t have gone over the limit
Hope this helps!!
Step-by-step explanation:
Ted's second job offers 49,024 and it is 3819 greater than the first job
The diameter is the cross section of the widest part of a sphere.