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1 year ago

Give the angle measure represented by 36.25 rotations counterclockwise

1 answer:

4 0

You could say 360(36.25)=13050° but all of the trigonometric relationships will still just obey that angle which is the fractional part because each rotation returns you to the positive x axis. The resulting angle is just .25(360)=90°.

So 13050° is the total angular rotation and 90° is the relative position with respect to the starting point.

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Complete the equation of the line through (-6,5)(−6,5)left parenthesis, minus, 6, comma, 5, right parenthesis and (-3,-3)(−3,−3)

goldenfox [79]

Answer:

The answer to your question is: y = -8/3 x - 11

Step-by-step explanation:

A (-6, 5)

B (-3, -3)

Process

1.- Find the slope of the line

Formula

$m = \frac{y2 - y1}{x2 - x1}$

Substitution

$m = \frac{-3 - 5}{-3 + 6}$

Simplifying

$m = \frac{- 8 }{3}$

2.- Find the equation of the line

Formula

( y - y1) = m(x - x1)

Substitution

( y + 3) = -8/3 (x + 3)

Simplifying

y + 3 = -8/3 x - 8

y = -8/3 x - 8 - 3

y = -8/3 x - 11

4 0

1 year ago

Solve the recurrence relation: hn = 5hn−1 − 6hn−2 − 4hn−3 + 8hn−4 with initial values h0 = 0, h1 = 1, h2 = 1, and h3 = 2 using (

musickatia [10]

(a) Suppose $h_n=r^n$ is a solution for this recurrence, with $r\neq0$ . Then

$r^n=5r^{n-1}-6r^{n-2}-4r^{n-3}+8r^{n-4}$
$\implies1=\dfrac5r-\dfrac6{r^2}-\dfrac4{r^3}+\dfrac8{r^4}$
$\implies r^4-5r^3+6r^2+4r-8=0$
$\implies (r-2)^3(r+1)=0\implies r=2,r=-1$

So we expect a general solution of the form

$h_n=c_1(-1)^n+(c_2+c_3n+c_4n^2)2^n$

With $h_0=0,h_1=1,h_2=1,h_3=2$ , we get four equations in four unknowns:

$\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}$

So the particular solution to the recurrence is

$h_n=-\dfrac8{27}(-1)^n+\left(\dfrac8{27}+\dfrac{7n}{72}-\dfrac{n^2}{24}\right)2^n$

(b) Let $G(x)=\displaystyle\sum_{n\ge0}h_nx^n$ be the generating function for $h_n$ . Multiply both sides of the recurrence by $x^n$ and sum over all $n\ge4$ .

$\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n$
$\displaystyle\sum_{n\ge4}h_nx^n=5x\sum_{n\ge3}h_nx^n-6x^2\sum_{n\ge2}h_nx^n-4x^3\sum_{n\ge1}h_nx^n+8x^4\sum_{n\ge0}h_nx^n$
$G(x)-h_0-h_1x-h_2x^2-h_3x^3=5x(G(x)-h_0-h_1x-h_2x^2)-6x^2(G(x)-h_0-h_1x)-4x^3(G(x)-h_0)+8x^4G(x)$
$G(x)-x-x^2-2x^3=5x(G(x)-x-x^2)-6x^2(G(x)-x)-4x^3G(x)+8x^4G(x)$
$(1-5x+6x^2+4x^3-8x^4)G(x)=x-4x^2+3x^3$
$G(x)=\dfrac{x-4x^2+3x^3}{1-5x+6x^2+4x^3-8x^4}$
$G(x)=\dfrac{17}{108}\dfrac1{1-2x}+\dfrac29\dfrac1{(1-2x)^2}-\dfrac1{12}\dfrac1{(1-2x)^3}-\dfrac8{27}\dfrac1{1+x}$

From here you would write each term as a power series (easy enough, since they're all geometric or derived from a geometric series), combine the series into one, and the solution to the recurrence will be the coefficient of $x^n$ , ideally matching the solution found in part (a).

3 0

2 years ago

A small ferryboat is 4.00 m wide and 6.00 m long. When a loaded truck pulls onto it, the boat sinks an additional 4.00 cm into t

Svet_ta [14]

Answer:

$F_w=9408\ N$

Weight of the truck=9408 N

Step-by-step explanation:

Boat is experiencing the buoyant force as it is in the water and is sinking

According to the force balance in y direction. As both is floating, two forces balance each other:

$F_b-F_w=0$

where:

$F_b$ is the buoyant force

$F_w$ is the weight=mg

$F_b=F_w$ Eq (1)

Buoyant force is equal to the mass of water displaced * gravitational acceleration.

$F_b=m_{water\ displaced}*g\\F_b=\rho Vg\\$

Taking density of water to be 1000 Kg/m^3

$F_b=1000*(6*4*4*10^{-2})*9.8\\F_b=9408 N$

From Eq(1):

$F_w=9408\ N$

Weight of the truck=9408 N

6 0

2 years ago

Marissa buys a licorice candy rope to share with some friends. The rope is 30 and two thirds inches long. She begins to cut the

adell [148]

9!
3.33 x 9 =(about)= 29.97

4 0

1 year ago

A hospital finds that 22% of its accounts are at least 1 month in arrears. A random sample of 425 accounts was taken. What is th

GenaCL600 [577]

Answer:

8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears

Step-by-step explanation:

For each account, there are only two possible outcomes. Either they are at least 1 month in arrears, or they are not. The probability of an account being at least 1 month in arrears is independent from other accounts. So the binomial probability distribution is used to solve this question.

However, we are working with a large sample. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .