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3 years ago

Which statement describes the graph of the system of equations below?

Mathematics

2 answers:

Orlov [11]3 years ago

8 0

It's no graph shown below

frozen [14]3 years ago

7 0

Answer: The lines intersect at (1.6,1.4)

Step-by-step explanation:

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A marketing research firm would like to survey undergraduate and graduate college students about whether or not they take out st

neonofarm [45]

Answer:

Objective Minimize 10x1 +15x2 + 12x3+18x4+15x5+ 21x6

The central total cost $ 17586 due to the number of central undergraduate students 1458 is very high.

The total minimum cost would $ 17586 +$260+ $300= $ 18146

Step-by-step explanation:

Let

X1 = # of undergraduate students from the East region,

X2 = # of graduate students from the East region,

X3 = # of undergraduate students from the Central region,

X4 = # of graduate students from the Central region,

X5 = # of undergraduate students from the West region, and

X6 = # of graduate students from the West region.

Then the cost functions are

y1= 10x1 +15x2

y2= 12x3+18x4

y3= 15x5+ 21x6

According to the given conditions

The constraints are

x1 +x2 + x3+x4+ x5+ x6 ≥ 1500------- A

15X2 +18X4+21X6 ≥ 400---------B

21X6 ≥ 100

X6 ≥ 100/21

X6 ≥ 4.76

Taking

X6= 5

10X1 ≤ 500

X1 ≤ 500/10

X1≤ 5

18X4 ≥ 75

X4 ≥ 75/18

X4 ≥ 4.167

Taking

X4= 5

Putting the values

15X5+ 21X6 ≥ 300

15X5+ 21(5) ≥ 300

15X5+ 105 ≥ 300

15X5 ≥ 300-105

15X5 ≥ 195

X5 ≥ 195/15

X5 ≥ 13

Putting value of X6 and X4 in B

15X2 +18X4+21X6 ≥ 400

15X2 +18(5)+21(5) ≥ 400

15X2 +195 ≥ 400

15X2 ≥ 400-195

15X2 ≥ 205

X2 ≥ 205/15

X2 ≥ 13.67

Taking X2= 14

Now putting the values in the cost equations to check whether the conditions are satisfied.

y1= 10x1 +15x2

y1= 10 (5) + 15(14)= 50 + 210= $ 260

y3= 15x5+ 21x6

y3= 15 (13) + 21(5)

y3= 195+105= $ 300

x1 +x2 + x3+x4+ x5+ x6 ≥ 1500

5+14+x3+5+13+5≥ 1500

x3≥ 1500-42

x3≥ 1458

y2= 12x3+18x4

y2= 12 (1458) + 18 (5)

y2= 17496 +90

y2= $ 17586

The cost can be minimized if the number of students from

Undergraduate Graduate

East Region X1≤ 5 X2 ≥ 13.67

Central X3≥ 1458 X4 ≥ 4.167

West X5 ≥ 13 X6 ≥ 4.76

This will result in the required number of students that is 1500

Constraints:

East Undergraduate must not be greater than 5

East Graduate must not be less than 13

Central Undergraduate must be greater than 1458

Central Graduate must be greater than 4

West Undergraduate must be greater than 13

West Graduate must be greater than 4

The central total cost $ 17586 due to the number of central undergraduate students 1458 is very high.

The east region has a least cost of $260 and west region has a cost of $300.

The total minimum cost would $ 17586 +$260+ $300= $ 18146

6 0

3 years ago

Simplify. Can you explain it also?

Doss [256]

Answer:

The answer is

Step-by-step explanation:

To solve the fraction reduce the fraction with d

That's we have

Next simplify the expression using the rules of indices to simplify the letters in the fraction

<u>For c </u>

Since they are dividing we subtract the exponents

We have

<u>For </u><u>e</u>

Substituting them into the expression we have

Reduce the fraction by 3

We have the final answer as

Hope this helps you

8 0

3 years ago

I'm in trouble. Please solve this two questions!!

melomori [17]

Answer:

.........................

4 0

3 years ago

Read 2 more answers

THERE ARE 3 BAGS OF APPLES WEIGHING A TOTAL OF 22 1/2 POUNDS. TWO OF THE BAGS WEIGH 6 3/8 POUNDS AND 3 1/4 POUNDS. HOW MUCH DOES

Mkey [24]

I wasn't going to click on this one, but the all-caps enthralled me and hypnotized me.

first off, let's change all the mixed fractions to "improper" and proceed from there, keep in mind that if we subtract the two bags' weight from the total, what's leftover is the 3rd bag's weight.

$\bf \stackrel{mixed}{22\frac{1}{2}}\implies \cfrac{22\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{45}{2}}
\\\\\\
\stackrel{mixed}{6\frac{3}{8}}\implies \cfrac{6\cdot 8+3}{8}\implies \stackrel{improper}{\cfrac{51}{8}}
\\\\\\
\stackrel{mixed}{3\frac{1}{4}}\implies \cfrac{3\cdot 4+1}{4}\implies \stackrel{improper}{\cfrac{13}{4}}\\\\
-------------------------------\\\\$

$\bf \stackrel{\textit{sum of the two bags}}{\cfrac{51}{8}+\cfrac{13}{4}}\impliedby \textit{our LCD is 8}\implies \cfrac{(1)51+(2)13}{8}
\\\\\\
\cfrac{51+26}{8}\implies \cfrac{77}{8}\\\\
-------------------------------\\\\
\stackrel{total}{\cfrac{45}{2}}~-~\stackrel{two~bags}{\cfrac{77}{8}}\impliedby \textit{our LCD is again 8}\implies \cfrac{(4)45-(1)77}{8}
\\\\\\
\cfrac{180~~-~~77}{8}\implies \cfrac{103}{8}\implies \stackrel{third~bag}{12\frac{7}{8}}$

3 0

3 years ago

Solve x2 + 8x − 3 = 0 using the completing-the-square method.

lisabon 2012 [21]

$(a+b)^2=a^2+2ab+b^2\\------------------\\\\x^2+8x-3=0\ \ \ |add\ 3\ to\ both\ sides\\\\x^2+2x\cdot4=3\ \ \ |add\ 4^2\ to\ both\ sides\\\\x^2+2x\cdot4+4^2=3+4^2\\\\(x+4)^2=3+16\\\\(x+4)^2=19\iff x+4=-\sqrt{19}\ or\ x+4=\sqrt{19}\\|subtract\ 4\ from\ both\ sides\\\\\boxed{x=-4-\sqrt{19}\ or\ x=-4+\sqrt{19}}$

3 0

3 years ago

Read 2 more answers