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Semmy [17]
3 years ago
11

Write an equation for a rational function with: Vertical asymptotes at x = -3 and x = -4 x-intercepts at x = 3 and x = 4 Horizon

tal asymptote at y = 10
Mathematics
1 answer:
sattari [20]3 years ago
8 0

Answer:

(10x ^ 2 - 70 * x +120) / (x ^ 2 + 7 * x +12)

Step-by-step explanation:

First a rational function is a function that is in the form a / b.

Knowing this, we proceed to calculate the asymptotes in x.

Vertical asymptotes

So that there is an asymptote in x, in a rational function the easiest thing is that the denominator is 0. Therefore, for the asymptotes x = -3 and x = -4, we make it 0 in both cases. That is to say:

x + 3 = 0

x + 4 = 0

both, in the denominator. For the moment it would be something like this:

 1 / (x + 3) * (x + 4)

X Intersections

Now, for intersections, what we should do equation 0, and that the solution of x us of the value we want, in this case x = 3 and x = 4. Because here the denominator does not influence because when equal to 0 to a rational function, the denominator becomes 0. Similarly, it is equal to 0 as in the previous case, only this time they would go in the numerator. It would be:

x - 3 = 0

x - 4 = 0

 Therefore, the function would go like this: (x - 3) * (x - 4) / (x +3) * (x + 4),

Operating the following we have:

(x ^ 2 -4 * x - 3 * x +12) / (x ^ 2 + 4 * x + 3 * x +12)

(x ^ 2 - 7 * x +12) / (x ^ 2 + 7 * x +12)

having asymptotes at x = -3 and x = -4, and intersection at (3.0) and (4.0), that is x = 3 and x = 4.

Horizontal asymptote.

Finally, in the case of the horizontal asymptote, which must be y = 10, what must be done is to divide all the terms by the highest degree term, in this case it is x ^ 2. It would be as follows:

(x ^ 2 / x ^ 2 - 7 * x / x ^ 2 + 12 / x ^ 2) / (x ^ 2 / x ^ 2 + 7 * x / x ^ 2 + 12 / x ^ 2)

Now, x ^ 2 / x ^ 2 = 1, and all the others have to be 0, therefore it would remain (1 + 0 + 0) / (1 + 0 + 0) = 1

So for the asymptote to be 10, the entire denominator must be multiplied by 10, thus being:

10 * (x ^ 2 - 7 * x +12) / (x ^ 2 + 7 * x +12)

(10x ^ 2 - 70 * x +120) / (x ^ 2 + 7 * x +12)

And this equation would already comply with everything required in the problem.

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Answer:

6 cm for both sides

Step-by-step explanation:

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3 years ago
Read 2 more answers
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xz_007 [3.2K]
Let's plug in each coordinate and see what we get:

(2, 0, 0)
-> 2A + 0 + 0 = D
-> 2A = D
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(0, 6, 0)
-> 0 + 6B + 0 = D
-> 6B = D
-> B = D/6

(0, 0, 5)
-> 0 + 0 + 5C = D
-> 5C = D
-> C = D/5

plugging this back into the equation:
(D/2)x + (D/6)y + (D/5)z = D    [then divide through by D]
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and since we have to have integer coefficients, 
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so we get
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Second question

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If you move three points in the y-direction, you'll reach to the  point A.
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5 0
3 years ago
Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x3 and y = x. (10 points)
USPshnik [31]
See the graph attached.

The midpoint rule states that you can calculate the area under a curve by using the formula:
M_{n} = \frac{b - a}{2} [ f(\frac{x_{0} + x_{1} }{2}) +  f(\frac{x_{1} + x_{2} }{2}) + ... +  f(\frac{x_{n-1} + x_{n} }{2})]

In your case:
a = 0
b = 1
n = 4
x₀ = 0
x₁ = 1/4
x₂ = 1/2
x₃ = 3/4
x₄ = 1

Therefore, you'll have:
M_{4} = \frac{1 - 0}{4} [ f(\frac{0 +  \frac{1}{4} }{2}) +  f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) +  f(\frac{\frac{1}{2} + \frac{3}{4} }{2}) + f(\frac{\frac{3}{4} + 1} {2})]
M_{4} = \frac{1}{4} [ f(\frac{1}{8}) +  f(\frac{3}{8}) +  f(\frac{5}{8}) + f(\frac{7}{8})]

Now, to evaluate your f(x), you need to look at the graph and notice that:
f(x) = x - x³

Therefore:
M_{4} = \frac{1}{4} [(\frac{1}{8} - (\frac{1}{8})^{3}) + (\frac{3}{8} - (\frac{3}{8})^{3}) + (\frac{5}{8} - (\frac{5}{8})^{3}) + (\frac{7}{8} - (\frac{7}{8})^{3})]

M_{4} = \frac{1}{4} [(\frac{1}{8} - \frac{1}{512}) + (\frac{3}{8} - \frac{27}{512}) + (\frac{5}{8} - \frac{125}{512}) + (\frac{7}{8} - \frac{343}{512})]

M₄ = 1/4 · (2 - 478/512)
     = 0.2666

Hence, the <span>area of the region bounded by y = x³ and y = x</span> is approximately 0.267 square units.

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3 years ago
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9966 [12]

Answer:

74

Step-by-step explanation:

6 0
3 years ago
Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were millio
JulijaS [17]

Answer:

The amount of oil was decreasing at 69300 barrels, yearly

Step-by-step explanation:

Given

Initial =1\ million

6\ years\ later = 500,000

Required

At what rate did oil decrease when 600000 barrels remain

To do this, we make use of the following notations

t = Time

A = Amount left in the well

So:

\frac{dA}{dt} = kA

Where k represents the constant of proportionality

\frac{dA}{dt} = kA

Multiply both sides by dt/A

\frac{dA}{dt} * \frac{dt}{A} = kA * \frac{dt}{A}

\frac{dA}{A}  = k\ dt

Integrate both sides

\int\ {\frac{dA}{A}  = \int\ {k\ dt}

ln\ A = kt + lnC

Make A, the subject

A = Ce^{kt}

t = 0\ when\ A =1\ million i.e. At initial

So, we have:

A = Ce^{kt}

1000000 = Ce^{k*0}

1000000 = Ce^{0}

1000000 = C*1

1000000 = C

C =1000000

Substitute C =1000000 in A = Ce^{kt}

A = 1000000e^{kt}

To solve for k;

6\ years\ later = 500,000

i.e.

t = 6\ A = 500000

So:

500000= 1000000e^{k*6}

Divide both sides by 1000000

0.5= e^{k*6}

Take natural logarithm (ln) of both sides

ln(0.5) = ln(e^{k*6})

ln(0.5) = k*6

Solve for k

k = \frac{ln(0.5)}{6}

k = \frac{-0.693}{6}

k = -0.1155

Recall that:

\frac{dA}{dt} = kA

Where

\frac{dA}{dt} = Rate

So, when

A = 600000

The rate is:

\frac{dA}{dt} = -0.1155 * 600000

\frac{dA}{dt} = -69300

<em>Hence, the amount of oil was decreasing at 69300 barrels, yearly</em>

7 0
2 years ago
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