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Phantasy [73]
4 years ago
5

I need help on this pleasssee!!

Mathematics
1 answer:
aev [14]4 years ago
8 0

Answer:

Option D. x=-3

Step-by-step explanation:

we have

f(x)=\frac{5x^{4}-7}{x+3}

we know that

The denominator cannot be equal to zero

therefore

To find the equation of the vertical asymptote equate the denominator to zero

so

(x+3)=0

x=-3 -----> equation of the vertical asymptote

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Which of the following definite integrals could be used to calculate the total area bounded by the graph of y = x^3 – 3x^2 + 2x
castortr0y [4]

Answer: The correct option is second, i.e. ,"2 times the integral from 0 to 1 of the quantity x cubed minus 3 times x squared plus 2 times x, dx".

Explanation:

The given equation is,

y=x^3-3x^2+2x

It can be written as,

f(x)=x^3-3x^2+2x

Find the zeros of the equation. Equation the function equal to 0.

0=x^3-3x^2+2x

x(x^2-3x+2)=0

x(x^2-2x-x+2)=0

x(x-2)(x-1)=0

So, the three zeros are 0, 1 and 2.

The graph of the equation is shown below.

From the given graph it is noticed that the enclosed by the curve and x- axis is lies between 0 to 2, but the area from 0 to 1 lies above the x-axis and area from 1 to 2 lies below the x-axis. So the function will be negative from 1 to 2.

The area enclosed by curve and x-axis is,

A=\int_{0}^{1}f(x)dx+\int_{1}^{2}[-f(x)]dx

A=\int_{0}^{1}f(x)dx-\int_{1}^{2}f(x)dx

From the graph it is noticed that the area from 0 to 1 is symmetric or same as area from 1 to 2. So the total area is the twice of area from 0 to 1.

A=2\int_{0}^{1}f(x)dx

A=2\int_{0}^{1}[x^3-3x^2+2x]dx

Therefore, The correct option is "2 times the integral from 0 to 1 of the quantity x cubed minus 3 times x squared plus 2 times x, dx".

3 0
3 years ago
Find the missing symbol: 1/6 ? 1/2<br>&lt;<br>&gt;<br>=<br>please help​
Amanda [17]

Answer:

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Step-by-step explanation:

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3 years ago
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Leya [2.2K]
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3 0
4 years ago
Can I please get some help
Aleksandr [31]

Answer:

a. 1.76

b. 2.5

Step-by-step explanation:

You have the following formula given in the exercise:

t=\frac{\sqrt{100-h}}{4}

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h=50

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This is:

t=\frac{\sqrt{100-50}}{4}\\\\t=1.76

b. When the object reaches the ground, the value of "h" is:

h=0

Therefore, substituting this value into the formula and evaluating, you get that time (in seconds) it takes to the object reaches the ground is:

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3 years ago
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Kitty [74]
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