Answer:
(14 - x) + (4·x - 22) = 3·x - 8
Answer:
y=2 and x=1
Step-by-step explanation:
2y=7y-14+4
2y-7y=-14+4
-5y=-10/-5
y=2
4x=6-4+2x
4x-2x=6-4
2x=2/2
x=1
Answer:
Profit = $18.8 per meal
Step-by-step explanation:
Chef ordered beef = 4 pounds
Cost of beef = $3.25 per pound
Total cost of beef = 4 × 3.25 = $13
Amount of potatoes ordered = 6 pounds
Cost of potatoes = $1.96 per pound
Total cost of potatoes = 6 × 1.96
= $11.76
Amount of carrots ordered = 5 pounds
Cost of carrots = $1.81 per pound
Total cost of carrots = 5 × 1.81 = $9.05
Total amount of beef, potatoes and carrots = $13 + $11.76 + $9.05
= $33.81
Number of meals prepared with material purchased = 28
Per meal cost = 
= $1.2075
Selling price of one meal = $20
Profit per meal = Selling price - Cost price
= 20 - 1.2075
= 18.7925
≈ $18.8
Therefore, he will make $18.8 for each meal.
Answer:
At first, we divide the parallelogram into two triangles by joining any two opposite vertices. These two triangles are exactly the same (congruent) and thus have equal areas. The area of the parallelogram is the summation of the individual areas of the two triangles. We drop a perpendicular from a vertex to its opposite side to get an expression for the height of the triangles. The area of the individual triangle is 12×base×height12×base×height .The area of the parallelogram being twice the area of the triangle, thus becomes after evaluation base×heightbase×height .
Complete step by step answer:
The parallelogram can be divided into two triangles by constructing a diagonal by joining any two opposite vertices.


In the above figure, ΔABDΔABD and ΔBCDΔBCDare the two such triangles. These two triangles have:
AB=CDAB=CD (as opposite sides of a parallelogram are equal)
AD=BCAD=BC (opposite sides of a parallelogram are equal)
BDBD is common
Thus, the two triangles are congruent to each other by SSS axiom of congruence. Since, the areas of two congruent triangles are equal,
⇒area(ΔABD)=area(ΔBCD)⇒area(ΔABD)=area(ΔBCD)
Now, we need to find the area of ΔABDΔABD . We draw a perpendicular from DD to the side ABAB and name it as DEDE . Thus, ΔABDΔABD is now a triangle with base ABAB and height DEDE .
Then, the area of the ΔABD
