Multiplying both sides of a rational equation by a variable expression introduces the possibility of extraneous solutions. Therefore, we must check the solutions against the set of restrictions. If a solution is a restriction, then it is not part of the domain and is extraneous.
Answer:
40-15-12= 13
Step-by-step explanation:
15 + 12 + x = 40
Solve for x
Jacob wants to build a rectangular enclosure for his animals. One side of the pen will be against the barn,so he needs no fence on that side.
Let w be the width of the enclosure (perpendicular to the barn) and let l be the length of the enclosure (parallel to the barn).
one side of the length is not counted for perimeter because one side of length will be against the barn.
Perimeter = 400 ft
Perimeter of rectangle = L + W + W
400 = L + 2W
L = 400 - 2W
Area = L * W
Replace L by 400 - 2W
A(W) = (400 - 2W) * W
Now we find out x coordinate of vertex to find the width that maximize the area
a= -2 and b = 400
The width w would maximize the area is w = 100ft
To find maximum area we plug in 100 for W in A(W)
the maximum area is 20,000 square feet
Answer: B. Exponential. There is a constant rate of decay or decrease.
The y-values decrease by 1/4 of the number that comes before every time.
The answer would be 16 S'mores and the limiting reactant would be the grahams.
(This is assuming that S'mores would need 2 grahams, 1 marshmallow and 1 chocolate piece.)
Limiting reactant would be the reactant that runs out first.
Let's take your problem into account and see what we have:
48 marshallows
32 grahams (2 x 16 per pack)
45 chocolate pieces (5 x 15 pieces per bar)
Since need 2 of the grahams per S'more then the maximum yield of the grahams is 16 S'mores.
The maximum yield of marshmallows is 48.
The maximum yield of chocolate is 45.
Since you cannot make S'mores without the grahams, then you can only make 16 S'mores before the grahams run out.
