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2 years ago

What conclusions can be made about the series ∞ 6 cos(πn) n n = 1 and the integral test?

1 answer:

7 0

Depending on how the function is set up, i presume that it is a sum (epsilon) function that is being referred to. Depending on if the function is increasing or decreasing on the interval (infinity to 6??), will determine if the function is convergent or not. When 1 is plugged into cos( $(\pi n)n$ it is positive and therefor converging.

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Which input value produces the same output value for the two functions on the graph? f(x) equals negative StartFraction 2 Over 3

alisha [4.7K]

Answer:

Step-by-step explanation:

Your problem statement tells you ....

f(x) passes through point (3, negative 1)
g(x) passes through point (3, negative 1)

This tells you the input value that produces the same output value for the two functions is 3, and that output value is -1.

8 0

2 years ago

Read 2 more answers

A parabolic satellite dish reflects signals to the dish’s focal point. An antenna designer analyzed signals transmitted to a sat

ivolga24 [154]

Answer:

$c=\frac{16}{39}$

Step-by-step explanation:

The probability density function is :

$f(x)=c(1-\frac{1}{16}x^{2})$

With 0 < x < 3

To be a valid probability density function :

$\int\limits^b_a {f(x)} \,dx=1$

Where a < x < b

And also

f(x) ≥ 0 for a < x < b

Applying this to the probability density function of the exercise :

$\int\limits^3_0 {c(1-\frac{1}{16}x^{2})} \, dx=1$

$c\int\limits^3_0 {(1-\frac{1}{16}x^{2})} \, dx=1$

$c(3-\frac{1}{16}\frac{3^{3}}{3})=1$

$c(\frac{39}{16})=1$

$c=\frac{16}{39}$

We can verify by replacing ''c'' in the original probability density function and integrating :

$\int\limits^3_0 {\frac{16}{39}(1-\frac{1}{16}x^{2})} \, dx=$

$=\frac{16}{39}.(3)-\frac{1}{39}.(\frac{3^{3}}{3})=\frac{16}{13}-\frac{3}{13}=\frac{13}{13}=1$

Also, f(x) ≥ 0 for 0 < x < 3

4 0

2 years ago

Fiona wrote the linear equation y = x – 5. when henry wrote his equation, they discovered that his equation had all the same sol

djverab [1.8K]

I would go with A but I am not 100% sure so you may want to check however you can.

6 0

2 years ago

Read 2 more answers

Choose the function whose graph is given by:

loris [4]

<h2>Answer:</h2>

$y=3sec\left(\frac{1}{2}x\right)$

<h2>Step-by-step explanation:</h2>

The graphs of $sec(x)$ can be obtained from the graph of the cosine function using the reciprocal identity, so:

$sec(x)=\frac{1}{cos(x)}$

But in this problem, the graph stands for the function:

$y=3sec\left(\frac{1}{2}x\right)$

Because the period is now 4π as indicated and for $x=0$ in the figure and this can be proven as follows:

$Period=\frac{2\pi}{\frac{1}{2}}=4\pi$

Also, $for \ x=1 \ then \ y=3$ as indicated in the figure and this can be proven as:

$y=3sec\left(\frac{1}{2}x\right) \\ \\ y=\frac{3}{cos(0.5x)} \\ \\ y=\frac{3}{cos(0.5(0))} \\ \\ y=\frac{3}{1}=3$

5 0

2 years ago

A customer borrowed $2000 and then a further $1000 both repayable in 12 months. What should he have saved if he had taken out on

Neporo4naja [7]

Answer:

a. $60

Step-by-step explanation:

We will use simple interest formula to solve our given problem.

$A=P(1+rt)$ , where

A= Amount after t years.

P= Principal amount.

r= Interest rate in decimal form.

t= Time in years.

Let us find amount of loans repayable after 12 months for taking two amounts of $2000 and $1000.

As $2000 and $1000 are less than 2500, so the rate of loan will be 10%.

$10\%=\frac{10}{100}=0.10$

12 months = 1 year.

$A=2000(1+0.10\times 1)$

$A=2000(1+0.10)$

$A=2000(1.10)$

$A=2200$

Now let us find amount repayable after 12 months for borrowing $1000.

$A=1000(1+0.10\times 1)$

$A=1000(1+0.10)$

$A=1000(1.10)$

$A=1100$

Adding these amounts we will get total repayable amount after 12 months for borrowing $2000 and $1000 separately.

$\text{Amount repayable for borrowing two separate amounts}=2200+1100=3300$

Now let us find repayable amount after 12 months for taking 1 loan. As $3000 is between $2501 and $7500, so rate of loan will be 8%.

$8\%=\frac{8}{100}=0.08$

$A=3000(1+0.08\times 1)$

$A=3000(1+0.08)$

$A=3000(1.08)$

$A=3240$

Now let us find difference between both repayable loan amounts.

$\text{Difference between both repayable loan amounts}=3300-3240$

$\text{Difference between both repayable loan amounts}=60$

Therefore, the customer should have saved $60, if he had taken out one loan for $3000 and option a is the correct choice.

6 0

2 years ago

Read 2 more answers