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3 years ago

If the graph f(x)= 9x^2+37x+41/3x+5 has an oblique asymptote at y=3x+k what is the value of k

Mathematics

2 answers:

ira [324]3 years ago

6 0

Answer:

$k=\frac{22}{3}$

Step-by-step explanation:

The oblique asymptote of

$f(x)=\frac{9x^2+37x+41}{3x+5}$ ,

We perform the long division as shown in the attachment.

The quotient is;

$3x+\frac{22}{3}$

Comparing to 3x+k

Hence the value of k is $\frac{22}{3}$

Archy [21]3 years ago

3 0

Answer:

$\frac{22}{3}$

Step-by-step explanation:

To find out oblique asymptote we divide the polynomials using long division

To find quotient divide the first term. then multiply the answer with 3x+5 and write it down. Subtract it from the top. Repeat the process till we get remainder.

$3x+\frac{22}{3}$

------------------------------

$3x+5$ $9x^2+37x+41$

$9x^2+15x$

-------------------------------------(Subtract)

$22x+41$

$22x+\frac{110}{3}$

------------------------------------(subtract)

$\frac{13}{3}$

Quotient is $3x+\frac{22}{3}$ that is our oblique asympotote

the value of k is 22/3

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weeeeeb [17]

Answer:

5.5%

Step-by-step explanation:

To find the interest rate, use the equation: I = prt

I = Interest = $6,600

p = principal = $15,000

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t = time = 8 years

Now plug the numbers into the equation.

6,600 = 15,000 * r * 8

6,600 = 120,000 * r

Now isolate the variable by dividing both sides of the equation by 120,000.

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Now change the rate to a percent by moving the decimal point 2 digits to the right.

5.5% = r OR r = 5.5%

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2 years ago

For the following hypothesis test, determine the null and alternative hypotheses. Also, classify the hypothesis test as two-tail

bagirrra123 [75]

Answer:

Option B:

$H_0: \mu = 22.1$

$H_a: \mu \neq 22.1$

Classification:

The hypothesis test is Two-tailed.

Step-by-step explanation:

The mean length of imprisonment for motor-vehicle theft offenders in this country is 22.1 months.

This means that the null hypothesis is that the mean is of 22.1 months, that is:

$H_0: \mu = 22.1$

A hypothesis test is to be performed to determine whether the mean length of imprisonment for motor-vehicle theft offenders in this city differs from the national mean of 22.1 months.

At the alternate hypothesis, we test if this mean is different of 22.1, that is:

$H_a: \mu \neq 22.1$

Which means that the answer is given by option b).

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We test if the mean is different from a value, which means that the hypothesis test is Two-tailed.

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3 years ago

An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t

USPshnik [31]

The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

$\displaystyle\int_a^bf(x)\,\mathrm dx$

Split up the interval $[a,b]$ into $n$ equal subintervals,

$[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]$

where $a=x_0$ and $b=x_n$ . Each subinterval has measure (width) $\dfrac{a-b}n$ .

Now denote the left- and right-endpoint approximations by $L$ and $R$ , respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are $\{x_0,x_1,\cdots,x_{n-1}\}$ . Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, $\{x_1,x_2,\cdots,x_n\}$ .

So, you have

$L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)$
$R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)$

Now let $T$ denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

$T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)$

Factoring out $\dfrac12$ and regrouping the terms, you have

$T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)$

which is equivalent to

$T=\dfrac12\left(L+R)$

and is the average of $L$ and $R$ .

So the trapezoidal approximation for your problem should be $\dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2$

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Answer:

Step-by-step explanation:

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