Which graph shows the feasible region for constraints x greater or equal to 1.

Mathematics

1 answer:

Nina [5.8K]3 years ago

3 0

Answer:

3rd option, (3,6) (5,10) (8,1) (14,1)

Step-by-step explanation:

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Select the best possible first step to solving the system by first eliminating the x variable.3x − 9y = 6 2x − 11y = 6?

musickatia [10]

$\left\{\begin{array}{ccc}3x-9y=6&|divide\ both\ sides\ by\ 3\\2x-11y=6&|divide\ both\ sides\ by\ (-2)\end{array}\right\\\\\underline{+ \left\{\begin{array}{ccc}x-3y=2\\-x+5.5y=-3\end{array}\right}\\.\ \ \ 2.5y=-1$

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3 years ago

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Solve the following equation for aa. Be sure to take into account whether a letter is capitalized or not.

Snezhnost [94]

Answer:

Since you state that capitalization is important, then we can see that it is not possible to solve the equation for F.

If that's correct, then the only point of this was that we need to pay attention to symbols when dealing with equations or expressions

Step-by-step explanation:

happy to help have a bless day or night:)

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3 years ago

What value of b will cause the system to have an infinite number of solutions?

irga5000 [103]

b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

<u>Solution: </u>

Need to calculate value of b so that given system of equations have an infinite number of solutions

$\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}$

Let us bring the equations in same form for sake of simplicity in comparison

$\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}$

Now we have two equations

$\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}$

Let us first see what is requirement for system of equations have an infinite number of solutions

If $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ are two equation

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ then the given system of equation has no infinitely many solutions.

In our case,

$\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}$