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Nuetrik [128]
2 years ago
6

Which of the following statements about feasible solutions to a linear programming problem is true?A. Min 4x + 3y + (2/3)z

Mathematics
1 answer:
Len [333]2 years ago
7 0

Answer:

The answer is "Option A"

Step-by-step explanation:

The valid linear programming language equation can be defined as follows:

Equation:

\Rightarrow \ Min\  4x + 3y + (\frac{2}{3})z

The description of a linear equation can be defined as follows:

It is an algebraic expression whereby each term contains a single exponent, and a single direction consists in the linear interpolation of the equation.

Formula:

\to \boxed{y= mx+c}

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If 20 pounds of beans sold for $1.20, what was the price per pound of beans?
zaharov [31]

Answer:

0.06 Cent

Step-by-step explanation:

Per Pound= Every one

so 1.20 divided by 20

= 0.06

So the price per pound of beans is 0.06 Cent

7 0
2 years ago
Read 2 more answers
What is -12 + (-4) + (-17) ??
Ludmilka [50]

Answer:

The answer is -33

Step-by-step explanation:

7 0
1 year ago
the population of masterton was 23 000 in 2012. Assume that Masterton population increases at a rate of 2% per year. Write an eq
Oksanka [162]
So, let's first focus on the first year:

the population then would be: 23000* (100+2)%

100% is the old population the 2% is the incease, so in total it's 102%, or 1.02 - the same number written differently.

So it will be 23000*1.02

After two years it will be 23000*1.02*1.02, and so on

so after x years it will be:

y=23000*1.02^{z}
8 0
3 years ago
The measures of the angles between the resultant and two applied forces are 26° and
erma4kov [3.2K]

Answer:

  48.8 lbs

Step-by-step explanation:

The forces can be modeled by a triangle with acute angles 20° and 26°, and obtuse angle 134° opposite a side of length 80. The larger component force will be opposite the angle 26°, and can be found using the Law of Sines:

  a/sin(A) = c/sin(C)

  x/sin(26°) = 80/sin(134°)

  x = 80sin(26°)/sin(134°) ≈ 48.753 . . . . pounds

The larger component force is about 48.8 pounds.

6 0
1 year ago
Vector u has a magnitude of 7 units and a direction angle of 330°. Vector v has magnitude of 8 units and a direction angle of 30
Dafna11 [192]
Keeping in mind that x = rcos(θ) and y = rsin(θ).

we know the magnitude "r" of U and V, as well as their angle θ, so let's get them in standard position form.

\bf u=
\begin{cases}
x=7cos(330^o)\\
\qquad 7\cdot \frac{\sqrt{3}}{2}\\
\qquad \frac{7\sqrt{3}}{2}\\
y=7sin(330^o)\\
\qquad 7\cdot -\frac{1}{2}\\
\qquad -\frac{7}{2}
\end{cases}\qquad \qquad v=
\begin{cases}
x=8cos(30^o)\\
\qquad 8\cdot \frac{\sqrt{3}}{2}\\
\qquad \frac{8\sqrt{3}}{2}\\
y=8sin(30^o)\\
\qquad 8\cdot \frac{1}{2}\\
\qquad 4
\end{cases}

\bf u+v\implies \left( \frac{7\sqrt{3}}{2},-\frac{7}{2} \right)+\left( \frac{8\sqrt{3}}{2},4 \right)\implies \left( \frac{7\sqrt{3}}{2}+\frac{8\sqrt{3}}{2}~~,~~ -\frac{7}{2}+4\right)
\\\\\\
\left(\stackrel{a}{\frac{15\sqrt{3}}{2}}~~,~~  \stackrel{b}{\frac{1}{2}}\right)\\\\
-------------------------------

\bf tan(\theta )=\cfrac{b}{a}\implies tan(\theta )=\cfrac{\frac{1}{2}}{\frac{15\sqrt{3}}{2}}\implies tan(\theta )=\cfrac{1}{15\sqrt{3}}
\\\\\\
\measuredangle \theta =tan^{-1}\left( \cfrac{1}{15\sqrt{3}} \right)\implies \measuredangle \theta \approx 2.20422750397203^o
8 0
2 years ago
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