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3 years ago

#6 please help! i will mark brainliest!

Mathematics

2 answers:

exis [7]3 years ago

6 0

Answer:

option 3

the cost of gasoline as a function of the number of gasoline as a function of the number of gallons purchased ( c = rg)

kramer3 years ago

5 0

Answer:

Choice 3

Step-by-step explanation:

You might be interested in

How do you find the rate of change

olga_2 [115]

If it's for a graph then I believe you just find the slope which you can do by taking two points and plugging them into this equation y-y
/
X-X

Example: (2, 4) and (6, 7) the first number is always x and the second is always y so if you plug those in the equation would look like this....

4-7
/
2-6

That would end up being
3
- ---- aka the rate of change
4

P.s. The slash and the four dotted lines together means division. Hope this helps

6 0

2 years ago

Compute: (1 2/5 +3.5÷1 1/4 )÷2 2/5 +3.4÷2 1 /8 −0.35

nikdorinn [45]

Answer:

Answer: 3

Step-by-step explanation:

Use BODMAS

Step 1: Open bracket (1 2/5 +3.5÷1 1/4 )

Convert mixed fractions into improper fractions

7/5 + 3.5 ÷ 5/4

Divide 3.5 by 5/4

7/5 + 2.8 = 4.2

Step 2: Carry out all divisions

Convert mixed fraction into improper fractions

4.2 ÷2 2/5 +3.4÷2 1 /8 −0.35

4.2 ÷ 12/5 + 3.4 ÷ 17/8 - 0.35

1.75 + 1.6 - 0.35

Step 3: Solve

1.75 + 1.6 - 0.35

3.35 - 0.35

Answer = 3

4 0

3 years ago

The Springer Dog Food Company makes dry dog food from two ingredients. The two ingredients (A and B) provide different amounts o

Nataly [62]

Answer:

a) Objective function (minimize cost):

$C=0.50A+0.20B$

Restrictions

Proteins per pound: $16A+8B\leq 12$

Vitamins per pound: $4A+8B\leq 6$

Non-negative values: $A,B\geq0$

b) Attached

c) The optimum solution (minimum cost) is 0 pounds of ingredient A and 0.75 pounds of ingredient B. The cost is $0.15 per ration.

d) The optimum solution changes. The cost is now 0 pounds of ingredient A and 0.625 pounds of ingredient B. The cost is $0.125 per ration.

Step-by-step explanation:

a) The LP formulation for this problem is:

Objective function (minimize cost):

$C=0.50A+0.20B$

Restrictions

Proteins per pound: $16A+8B\leq 12$

Vitamins per pound: $4A+8B\leq 6$

Non-negative values: $A,B\geq0$

b) The feasible region is attached.

c) We have 3 corner points. In one of them lies the optimal solution.

Corner A=0 B=0.75

$C=0.50*0+0.20*0.75=0.15$

Corner A=0.5 B=0.5

$C=0.50*0.5+0.20*0.5=0.35$

Corner A=0.75 B=0

$C=0.50*0.75+0.20*0=0.375$

The optimum solution (minimum cost) is 0 pounds of ingredient A and 0.75 pounds of ingredient B. The cost is $0.15 per ration.

d) If the company requires only 5 units of vitamins per pound rather than 6, one of the restrictions change.

The feasible region changes two of its three corners:

Corner A=0 B=0.625

$C=0.50*0+0.20*0.625=0.125$

Corner A=0.583 B=0.333

$C=0.50*0.583+0.20*0.333=0.358$

Corner A=0.75 B=0

$C=0.50*0.75+0.20*0=0.375$

The optimum solution changes. The cost is now 0 pounds of ingredient A and 0.625 pounds of ingredient B. The cost is $0.125 per ration.

3 0

3 years ago

Sum of all natural numbers from100to300 which are divisible by 4and 5 is

My name is Ann [436]

If $x$ is a number that is both divisible by 4 and 5, then

$\begin{cases}x\equiv0\pmod4\\x\equiv0\pmod5\end{cases}$

4 and 5 are coprime, so we can use the Chinese remainder theorem to solve this system and find that $x=20n$ is a solution to the system, where $n$ is any integer. Simply put, any multiple of 20 fits the bill.

Now, there are 11 numbers between 100 and 300 that are divisible by 20 (100, 120, 140, and so on). We have $20n=100$ when $n=5$ , so the sum we want to compute is