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Delvig [45]
2 years ago
6

Plz HELP...(MATH)........

Mathematics
2 answers:
frozen [14]2 years ago
4 0

Answer:

36 times

Step-by-step explanation:

Because half the marbles are blue, you can predict that about half of the 72 times you will choose the blue marble, or 36 times.

elena-s [515]2 years ago
4 0
I was so late when I got to this question I could of helped earlier:/
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when conducting a perpendicular line through point of a line, how can you verify that the line contested is perpendicular?
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7 0
3 years ago
1 Point
alexandr1967 [171]

Answer:

Option C

Step-by-step explanation:

The given inequality is

5x-1 < 19 and -3-x+1\le1

Group similar terms to get:

5x < 19+1 and -x\le1-1+3

5x < 20 and -x\le3

Solve for x to get:

x < 4 and  x\ge -3

The correct choice is C

4 0
3 years ago
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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
Marrrta [24]

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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