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3 years ago

What is the area of the trapezoid shown? The figure is not drawn to scale.

Mathematics

1 answer:

guapka [62]3 years ago

8 0

Answer:

284.3

Step-by-step explanation:

Trapezoid Formula:

A=a+b/2*h

plug in the numbers

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

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

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

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

5 0

3 years ago

16/4+56-[3+4+1]=<br><br> WHATS THE ANSWER

kifflom [539]

First do the parentheses:
16/4+56-8
Then divide:
4+56-8
Then add and subtract from left to right:
60-8
52
Hope this helps.

5 0

3 years ago

Read 2 more answers

What’s the length of AG?

Ivenika [448]

Answer:

Explanation:
10^2 + 5^2 = c^2
100 + 25 = 125^2
sqrt of 125^2 = 11.2 (rounded-off)

4 0

3 years ago

Applications 24. A rocket is launched into the air. Its height in feet, after x seconds, is given by the equation The starting h

SVEN [57.7K]

The given function is

$h(x)=-16x^2+300x+20$

According to this function, the starting height of the rocket is 20 feet because that's the initial condition of the problem stated by the independent term.

Additionally, we find the maximum height by calculating the vertex of the function V(h,k).

$h=-\frac{b}{2a}$

Where a = -16 and b = 300.

$\begin{gathered} h=-\frac{300}{2(-16)}=\frac{300}{32}=\frac{150}{16}=\frac{75}{8} \\ h=9.375 \end{gathered}$

Then, we find k by evaluating the function

$\begin{gathered} k=-16(9.375)^2+300(9.375)+20 \\ k=-1406.25+2812.5+20=1426.25 \end{gathered}$

Hence, the maximum height is 1426.25 feet.

At last, to know the time need to hit the ground, we just use h=9.375 and we multiply it by 2

$t=2\cdot9.375=18.75$

Hence, the rocket hits the ground after 18.75 seconds.

6 0

1 year ago

A high school athletic department bought 40 soccer uniforms at a cost of 3,000. After soccer season they returned some of the un

zmey [24]

$3000/$40
$75 each
$75 - $40 = $35 (difference)

5 0

3 years ago