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notka56 [123]
3 years ago
15

What are some limitations of the linnaean classification system?

Mathematics
1 answer:
s2008m [1.1K]3 years ago
3 0
In his Imperium Naturae<span>, Linnaeus established three kingdoms, namely </span>Regnum Animale<span>, </span>Regnum Vegetabile<span> and </span>Regnum Lapideum<span>. This approach, the Animal, Vegetable and Mineral Kingdoms, survives today in the popular mind, notably in the form of the parlour game question: "Is it </span>animal, vegetable or mineral<span>?". The work of Linnaeus had a huge impact on science; it was indispensable as a foundation for </span>biological nomenclature<span>, now regulated by the </span>nomenclature codes<span>.</span><span>
But the system is based solely on physical characteristics, due to not having the proper technology to investigate the molecular data of animals</span>
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The height of a triangle is 4 ft less than the base x. The area is 126 ft2. Find the dimensions of the triangle.
valina [46]

Answer:

Base = x = 18ft

Height = x - 4 = 14ft

Step-by-step explanation:

The formula for the area of A the triangle is given by the following equation:

1) A = (base * heigth)/(2)

From this problem, we know that

base = x and that the height is 4ft less than the base, so height = base - 4 = x - 4.

We also know that the area is 126ft², so A = 126 ft². Now we can replace these informations in equation 1).

126 = (x(x-4))/2

x² - 4x = 252

x² -4x - 252 = 0

The roots of this equation are x = 18 or x = -14. But we know that a triangle cannot have negative dimensions, so we use x = 18.

The base is x, so base = 18ft.

Height is 4ft less than the base, so height equals 18-4 = 14ft

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3 years ago
Learn with an example
liubo4ka [24]

Answer:12

Step-by-step explanation:

36:3=12

6 0
3 years ago
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Solve the following differential equation: y" + y' = 8x^2
PtichkaEL [24]

Answer:

y=y_p+y_h = \frac{8}{3}x^3-8x^2+16x+ C_1 + C_2e^{-x}

Step-by-step explanation:

Let's  find a particular solution. We need a function of the form y= ax^3+bx^2+cx+d such that

y'= 3ax^2+2bx+c and

y''= 6ax+2b

y'+y''= 3ax^2+2bx+c+6ax+2b = 3ax^2+x(2b+6a)+(c+2b) = 8x^2

then, 3a= 8, 2b+6a =0 and c+2b = 0. With the first equation we obtain

a =  8/3 and replacing in the second equation 2b+6(8/3) = 2b + 16 = 0. Then, b = -8. Finally, c = -2(-8) = 16.

So, our particular solution is  y_p= \frac{8}{3}x^3-8x^2+16x.

Now, let's find the solution y_p of the homogeneus equation y''+y'=0 with the method of constants coefficients. Let y=e^{\lambda x}

y'=\lambda e^{\lambda x}

y''=\lambda^2e^{\lambda x}

then \lambda e^{\lambda x}+\lambda^2 e^{\lambda x} = 0

e^{\lambda x}(\lambda +\lambda^2)= 0

(\lambda +\lambda^2)= 0

\lambda (1+\lambda)= 0

\lambda =0 and \lambda)= -1.

So, y_h = C_1 + C_2e^{-x} and the solution is

y=y_p+y_h =\frac{8}{3}x^3-8x^2+16x+ C_1 + C_2e^{-x}.

5 0
3 years ago
Is 5.4 irrational or rational
salantis [7]
Irrational................................
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3 years ago
EXTRA POINTS<br> x/4 ÷ 2/3 = 1/5 solve for x
nikitadnepr [17]

Answer:

x=40

Step-by-step explanation:

first you simplify both sides of the equation, then you isolate the variable

40/4= 10

10 times 3/2 = 15

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