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DIA [1.3K]
3 years ago
7

The length of a rectangle is the width minus 3 units. The area of the rectangle is 40 units. What is the width, in units, of the

rectangle?
Mathematics
1 answer:
Eva8 [605]3 years ago
6 0

The width of rectangle is 8 units.

Step-by-step explanation:

Given,

Area of rectangle = 40 units

Width = w

Length = w-3

Area = Length * Width

40=(w-3)*w\\40=w^2-3w\\w^2-3w=40\\w^2-3w-40=0

Factorizing the equation

w^2-8w+5w-40=0\\w(w-8)+5(w-8)=0\\(w-8)(w+5)=0

Either,

w-8=0         => w=8

Or,

w+5=0        =>w= -5

As width cannot be negative, therefore

Width of rectangle = 8 units

The width of rectangle is 8 units.

Keywords: area, rectangle

Learn more about rectangles at:

  • brainly.com/question/10538663
  • brainly.com/question/10557275

#LearnwithBrainly

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Which is the standard form of the equation of the parabola that has a vertex of (3, 1) and a directric of x = -2?
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Check the picture below.  So,the parabola looks like so, notice the distance "p". since the parabola is opening to the right, then "p" is positive, thus is 5.

\bf \textit{parabola vertex form with focus point distance}\\\\
\begin{array}{llll}
\boxed{(y-{{ k}})^2=4{{ p}}(x-{{ h}})} \\\\
(x-{{ h}})^2=4{{ p}}(y-{{ k}}) \\
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\

\bf -------------------------------\\\\
\begin{cases}
h=3\\
k=1\\
p=5
\end{cases}\implies (y-1)^2=4(5)(x-3)\implies (y-1)^2=20(x-3)
\\\\\\
\cfrac{1}{20}(y-1)^2=x-3\implies \boxed{\cfrac{1}{20}(y-1)^2+3=x}

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n many population growth problems, there is an upper limit beyond which the population cannot grow. Many scientists agree that t
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Answer:

\frac{dP}{dt} = rP(1 - \frac{P}{K}) = 0.017P(1 - \frac{P}{16})

Step-by-step explanation:

The logistic function of population growth, that is, the solution of the differential equation is as follows:

P(t) = \frac{KP_{0}e^{rt}}{K + P_{0}(e^{rt} - 1)}

We use this equation to find the value of r.

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K = 16, P_{0} = 2, P(50) = 4

So we find the value of r.

P(t) = \frac{KP_{0}e^{rt}}{K + P_{0}(e^{rt} - 1)}

4 = \frac{16*2e^{50r}}{16 + 2*(e^{50r} - 1)}

4 = \frac{32e^{50r}}{14 + 2e^{50r}}

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