Answer:
How many drinks should be sold to get a maximal profit? 468
Sales of the first one = 345 cups
Sales of the second one = 123 cups
Step-by-step explanation:
maximize 1.2F + 0.7S
where:
F = first type of drink
S = second type of drink
constraints:
sugar ⇒ 3F + 10S ≤ 3000
juice ⇒ 9F + 4S ≤ 3600
coffee ⇒ 4F + 5S ≤ 2000
using solver the maximum profit is $500.10
and the optimal solution is 345F + 123S
<em>AC bisects ∠BAD, => ∠BAC=∠CAD ..... (1)</em>
<em>thus in ΔABC and ΔADC, ∠ABC=∠ADC (given), </em>
<em> ∠BAC=∠CAD [from (1)],</em>
<em>AC (opposite side side of ∠ABC) = AC (opposite side side of ∠ADC), the common side between ΔABC and ΔADC</em>
<em>Hence, by AAS axiom, ΔABC ≅ ΔADC,</em>
<em>Therefore, BC (opposite side side of ∠BAC) = DC (opposite side side of ∠CAD), since (1)</em>
<em />
Hence, BC=DC proved.
N= 1 a this is the answer hope i helped
- _ + _
r pr
Let's solve your equation step-by-step.
5=(−4)(8)+b
Step 1: Simplify both sides of the equation.
5=(−4)(8)+b
5=−32+b
5=b−32
Step 2: Flip the equation.
b−32=5
Step 3: Add 32 to both sides.
b−32+32=5+32
b=37
3/4, 7/16, 5/8.....LCD = 16
3/4 = 12/16
7/16
5/8 = 10/16
