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katen-ka-za [31]
2 years ago
5

A, E and B lie on a straight line segment.

Mathematics
1 answer:
Jobisdone [24]2 years ago
4 0

The length of line BE on the given line segment is 3.5 m.

The given parameters:

  • <em>Length BD = 3 m</em>
  • <em>Length CD = 15 m</em>
  • <em>Length AB = 21 m</em>

<em />

The sketch of the given line segments is presented in the diagram uploaded.

The length of BE is determined by applying similar triangle theorem as follows;

\frac{CB}{AB} = \frac{BD}{EB} \\\\\frac{15 + 3}{21} = \frac{3}{EB} \\\\18EB = 63 \\\\EB = \frac{63}{18} \\\\EB = 3.5 \ m

Thus, the length of line BE on the given line segment is 3.5 m.

Learn more about similar triangle here: brainly.com/question/11899908

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Find the critical numbers of the function f(x) = x6(x − 2)5.x = (smallest value)x = x = (largest value)(b) What does the Second
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Answer:

a) x=0, x=\frac{12}{11}, x=2 \: b) The 2nd Derivative test shows us the change of sign and concavity at some point. c) At which point the concavity changes or not. This is only possible with the 2nd derivative test.

Step-by-step explanation:

a) To find the critical numbers, or critical points of:

f(x)=x^{6}(x-2)^{5}

1) The procedure is to calculate the 1st derivative of this function. Notice that in this case, we'll apply the <em>Product Rule</em> since there is a product of two functions.

f(x)=x^{6}(x-2)^{5}\Rightarrow f'(x)=(f*g)'(x)\\=f'g+fg'\Rightarrow (fg)'(x)=6x^{5}(x-2)^{5}+5x^{6}(x-2)^{4} \Rightarrow 6x^{5}(x-2)^{5}+5x^{6}(x-2)^{4}=0\\f'(x)=6x^{5}(x-2)^{5}+5x^{6}(x-2)^{4}

2) After that, set this an equation then find the values for x.

x^{5}(x-2)^{4}[6(x-2)+5x]=0\Rightarrow x^{5}(x-2)^{4}[11x-12]=0\Rightarrow x_{1}=0\\(x-2)^{4}=0\Rightarrow \sqrt[4]{(x-2)}=\sqrt[4]{0}\Rightarrow x-2=0\Rightarrow x_{2}=2\\(11x-12)=0\Rightarrow x_{3}=\frac{12}{11}

x=0\:(smallest\:value)\:x_{3}=\frac{12}{11}\:x=2 (largest value)

b) The Second Derivative Test helps us to check the sign of given critical numbers.

Rewriting f'(x) factorizing:

f'(x)=(11x-12)(x-2)^4x^{5}

Applying product Rule to find the 2nd Derivative, similarly to 1st derivative:

f''(x)>0 \Rightarrow Concavity\: Up\\\\f''(x)

f''(x)=11\left(x-2\right)^4x^5+4\left(x-2\right)^3x^5\left(11x-12\right)+5\left(x-2\right)^4x^4\left(11x-12\right)\\f''(x)=10\left(x-2\right)^3x^4\left(11x^2-24x+12\right)

1) Setting this to zero, as an equation:

10\left(x-2\right)^3x^4\left(11x^2-24x+12\right)=0\\\\

10\left(x-2\right)^3x^4\left(11x^2-24x+12\right)=0\\(x-2)^{3}=0 \Rightarrow x_1=2\\x^{4}=0 \therefore x_2=0\\11x^{2}-24x+12=0 \Rightarrow x_3=\frac{12+2\sqrt{3}}{11}\:,x_4=\frac{12-2\sqrt{3}}{11}\cong 0.78

2) Now, let's define which is the inflection point, the domain is as a polynomial function:

D=(-\infty

Looking at the graph.

Plugging these inflection points in the original equationf(x)=x^{6}(x-2)^{5} to get y coordinate:

We have as Inflection Points and their respective y coordinates (Converting to approximate decimal numbers)

(1.09,-1.05) Inflection Point and Local Minimum

(2,0) Inflection Point and Saddle Point

(0,0) Inflection Point Local Maximum

(Check the graph)

c) At which point the concavity changes or not. This is only possible with the 2nd derivative test.

At

x=\frac{12}{11}\cong1.09 Local Minimum

At\:x=0,\:Local \:Maximum

At\:x=2, \:neither\:a\:minimum\:nor\:a\:maximum (Saddle Point)

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