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Alecsey [184]
2 years ago
14

Find the largest degree of x that can be factored out of all terms 9x and 45​

Mathematics
1 answer:
stiv31 [10]2 years ago
8 0

Step-by-step explanation:

x cannot be factored out of these two monomials because x is not present in both of them.

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You are taking hang-gliding lessons. The cost of the first lesson is one and a half times the cost of each additional lesson. Yo
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So, givens: total lesson cost of $260, total lessons taken are 6, and the first lesson costs 1.5 (or 3/2) as much as the additional lessons.

First thing to do is to figure out how many additional lessons are in that, which are 5.

Then you can make a 1 variable equation with the information you have. I’m using x as my variable.

260= 3/2x + 5x

Combine like terms.

260 = 13/2x

Divide both sides by 13/2 (treat it as a fraction, and if your calculator cannot make fractions, then decimal might help for this. 13/2=6.5)

X=40
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Select the statement that describes the expression 7+2n
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State the linear programming problem in mathematical terms, identifying the objective function and the constraints. A firm makes
Sedbober [7]

Answer:

Maximum profit at (3,0) is $27.

Step-by-step explanation:

Let  quantity of  products A=x

Quantity  of products B=y

Product A takes time on machine L=2 hours

Product A takes time on machine M=2 hours

Product B takes time on machineL= 4 hours

Product B takes time on machine M=3 hours

Machine L can used total time= 8hours

Machine M can used total time= 6hours

Profit on product A= $9

Profit on product B=$7

According to question

Objective function Z=9x+7y

Constraints:

2x+3y\leq 6

2x+4y\leq 8

Where x\geq 0, y\geq 0

I equation 2x+3y\leq 6

I equation in inequality change into equality we get

2x+3y=6

Put x=0 then we get

y=2

If we put y=0 then we get

x= 3

Therefore , we get two points A (0,2) and B (3,0) and plot the graph for equation I

Now put x=0 and y=0 in I equation in inequality

Then we get 0\leq 6

Hence, this equation is true then shaded regoin is  below the line .

Similarly , for II equation

First change inequality equation into equality equation

we get 2x+4y=8

Put x= 0 then we get

y=2

Put y=0 Then we get

x=4

Therefore, we get two points C(0,2)a nd D(4,0) and plot the graph for equation II

Point  A and C are same

Put x=0 and y=0 in the in inequality equation II then we get

0\leq 8

Hence, this equation is true .Therefore, the shaded region is below the line.

By graph we can see both line intersect at the points A(0,2)

The feasible region is AOBA and bounded.

To find the value of objective function on points

A (0,2), O(0,0) and B(3,0)

Put A(0,2)

Z= 9\times 0+7\times 2=14

At point O(0,0)

Z=0

At point B(3,0)

Z=9\times3+7\times0=27

Hence maximum value of z= 27 at point B(3,0)

Therefore, the maximum profit is $27.

6 0
3 years ago
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