We are required to find the total milligrams if medicine the hospital can get from the three companies.
The total quantity of medicine the hospital can get from the three companies is 2.75 milligrams
There are there companies:
Company A
Company B
Company C
Company A = 1.3 milligrams
Company A is 0.9 milligrams more than Company C's
A = 0.9 + C
1.3 = 0.9 + C
1.3 - 0.9 = C
0.4 = C
C = 0.4 milligrams
Company A is also twice the difference between the weight of
Company B's supply and Company C's
A = 2(B - C)
1.3 = 2(B - 0.4)
1.3 = 2B - 0.8
1.3 + 0.8 = 2B
2.1 = 2B
Divide both sides by 2
B = 2.1/2
B = 1.05 milligrams
Therefore,
A + B + C
= 1.3 + 1.05 + 0.4
= 2.75 milligrams
Read more:
brainly.com/question/24562139
EXPLANATION:
STEPS:
1.First we have to find the partial distances in the two equations given and then find the total, but to do so we must find the value of x, then replace it in each of the equations and finally the sum will give us the total of all the magnitudes.
2.
Answer:
B. 30°
Step-by-step explanation:
Answer:
the second one that is 3y+1
Answer:
Point Critical point
Q (2,0) local minimum
R (-2,1) Saddle
S (2,-1) local maximum
T ( -2,-1) Saddle
O ( -2,0) Saddle
Step-by-step explanation: INCOMPLETE ANSWER INFORMATION ABOUT THE POINTS ARE RARE
f(x,y) = x³ +y⁴ - 6x -2y² +3
df/dx = f´(x) = 3x² -6x
df/dxdx = f´´(xx) = 6x
df/dy = f´(y) = -4y
df/dydy = 4
df/dydx = df/dxdy = 0
df/dydy = f´´(yy)
D = [ df/dxdx *df/dydy] - [df/dydx]²
D = (6x)*4 - 0
D = 6*2*4 D > 0 and the second derivative on x is 6*2 = 12
so D > 0 and df/dxdx >0 there is a local minimum in P
Q(2,1)
D = (6*2)*4 D>0 and second derivative on x is 6*2
The same condition there is a minimum in Q
R ( -2,1)
D = 6*(-2)*4 = -48 D< 0 there is a saddle point in R
S (2,-1)
D = 6*2*4 = 48 D > 0 and df/dxdx = 6*-1 = -6
There is a maximum in S
T ( -2,-1)
D = 6*(-2)*(4) = -48 D<0 there is a saddle point in T
O ( -2,0)
D = 6*(-2)*4 = -48 D<0 there is a saddle point in O
