Ask question

Ask question

All categories

Natasha2012 [34]

1 year ago

13

ABCD is a rectangle and A and B are the points (4,2) and (2,8) respectively. Given that the equation of AC is y=x-2, find a) Equ

ation of BC b) coordinates of B and C

1 answer:

Rudik [331]1 year ago

8 0

A(4,2), B(2,8)

AC: y=x-2

Point point form for a line through (a,b) and (c,d) is (c-a)(y-b)=(d-b)(x-a)

AB is (2 - 4)(y - 2) = (8-2)(x-4)

-2(y - 2)=6(x-4)

y - 2 = -3(x-4)

y = -3x + 14

BC is perpendicular through B, so slope 1/3 and we calculate the constant as y-(1/3)x:

y = (1/3) x + (8 - (1/3)(2) ) = (1/3) x + 22/3

a)Answer: BC is y = (1/3) x + 22/3

C is the meet of AC and BC,

x - 2 = (1/3) x + 22/3

3x - 6 = x + 22

2x = 28

x = 14

y = x-2 = 12

Check: y = (1/3)x + 22/3 = (14+22)/3 = 36/3 = 12 good

C(14,12)

The remaining corner is D. We have A-B=D-C or

D = A+C-B+C = (4,2)+(14,12)-(2,8) = (16, 6)

b)Answer: B(2,8) [given, but asked for] C(14,12), D(16,6)

You might be interested in

There is a 0.9991 probability that a randomly selected 31-year-old male lives through the year. A life insurance company charge

jarptica [38.1K]

Answer:

a) Monetary values corresponding to the two events are:

-In case of surviving the year = -166$

-In case of a death in the year = 89834$

b) Expected value of the purchasing the insurance is -85 $

c) Yes, insurance company can make a profit with this policy.

Step-by-step explanation:

<em>a)</em> The man need to pay 166$ first to enroll the insurance policy. If he survives within a year, he will lose 166$. Otherwise, if he dies within a year he will profit 89834$.

<em>b)</em> Expected value of the purchasing the insurance as following:

<u>-In case of surviving the year: </u>

Value: -166$

Probability: 0,9991

<u>-In case of death in a year </u>

Value: 89834$

Probability: 0,0009

Expected value is E(x) = -166×0,9991 + 89834×0,0009 = -85 $

<em>c)</em> Lets consider that 10000 different 31 year old man enrolled to this insurance policy. According to probability of death, 9 out of 10000 man expected to be dead within the year. Therefore, company need to pay 9*90000 = 810000$ to their costumers. But, company will collect 10000*166=1660000$ from their costumers in the beginning of the year

So, it is expected that company is going to profit 1660000-810000=850000$ per year.

4 0

2 years ago

The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillat

Greeley [361]

Answer:

1) $L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

2) $T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

3) $T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

Step-by-step explanation:

Part 1

For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.

$L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

Part 2

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

Part 3

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

Replacing we got:

$T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

8 0

2 years ago

On a piece of paper, graph y< x-3. Then determine which answer matches<br> the graph you drew

Aneli [31]

Answer:

The graph is shown below.

Step-by-step explanation:

Given:

The inequality of a line to graph is given as:

$y$

In order to graph it, we first make the 'inequality' sign to 'equal to' sign. This gives,

$y=x-3$

Now, we plot this line on a graph. The given line is of the form:

$y=mx+b$ Where, 'm' is the slope and 'b' is the y-intercept.

So, for the line $y=x-3$ , $m=1,b=-3$

The y-intercept is at (0, -3).

In order to draw the line correctly we find another point. Let the 'y' value be 0.

Now, $0=x-3\\x=3$

So, the point is (3, 0).

Now, we mark these points and draw a line passing through these two points.

Now, consider the line inequality $y$ . The 'y' value is less than $x-3$ . So, the solution region will be region below the line and excluding all the points on the line. So, we draw a broken line and shade the region below it.

The graph is shown below.

5 0

2 years ago

Dee's credit card has an APR of 17% calculated on the previous monthly balance, and Dee makes a payment of $50 every month. Her

Bess [88]

Where is the table..? You didn't provide any pictures..

8 0

2 years ago

Read 2 more answers

Power series of y''+x^2y'-xy=0

Ray Of Light [21]

Assuming we're looking for a power series solution centered around $x=0$ , take

$y=\displaystyle\sum_{n\ge0}a_nx^n$
$y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}$
$y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}$

Substituting into the ODE yields

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}+\sum_{n\ge1}na_nx^{n+1}-\sum_{n\ge0}a_nx^{n+1}=0$

The first series starts with a constant term; the second series starts at $x^2$ ; the last starts at $x^1$ . So, extract the first two terms from the first series, and the first term from the last series so that each new series starts with a $x^2$ term. We have

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}=2a_2+6a_3x+\sum_{n\ge4}n(n-1)a_nx^{n-2}$

$\displaystyle\sum_{n\ge0}a_nx^{n+1}=a_0x+\sum_{n\ge1}a_nx^{n+1}$

Re-index the first sum to have it start at $n=1$ (to match the the other two sums):

$\displaystyle\sum_{n\ge4}n(n-1)a_nx^{n-2}=\sum_{n\ge1}(n+3)(n+2)a_{n+3}x^{n+1}$

So now the ODE is

$\displaystyle\left(2a_2+6a_3x+\sum_{n\ge1}(n+3)(n+2)a_{n+3}x^{n+1}\right)+\sum_{n\ge1}na_nx^{n+1}-\left(a_0x+\sum_{n\ge1}a_nx^{n+1}\right)=0$

Consolidate into one series starting $n=1$ :

$\displaystyle2a_2+(6a_3-a_0)x+\sum_{n\ge1}\bigg[(n+3)(n+2)a_{n+3}+(n-1)a_n\bigg]x^{n+1}=0$

Suppose we're given initial conditions $y(0)=a_0$ and $y'(0)=a_1$ (which follow from setting $x=0$ in the power series representations for $y$ and $y'$ , respectively). From the above equation it follows that

$\begin{cases}2a_2=0\\6a_3-a_0=0\\(n+3)(n+2)a_{n+3}+(n-1)a_n=0&\text{for }n\ge2\end{cases}$

Let's first consider what happens when $n=3k-2$ , i.e. $n\in\{1,4,7,10,\ldots\}$ . The recurrence relation tells us that

$a_4=-\dfrac{1-1}{(1+3)(1+2)}a_1=0\implies a_7=0\implies a_{10}=0$

and so on, so that $a_{3k-2}=0$ except for when $k=1$ .

Now let's consider $n=3k-1$ , or $n\in\{2,5,8,11,\ldots\}$ . We know that $a_2=0$ , and from the recurrence it follows that $a_{3k-1}=0$ for all $k$ .

Finally, take $n=3k$ , or $n\in\{0,3,6,9,\ldots\}$ . We have a solution for $a_3$ in terms of $a_0$ , so the next few terms ( $k=2,3,4$ ) according to the recurrence would be

$a_6=-\dfrac2{6\cdot5}a_3=-\dfrac2{6\cdot5\cdot3\cdot2}a_0=-\dfrac{a_0}{6\cdot3\cdot5}$
$a_9=-\dfrac5{9\cdot8}a_6=\dfrac{a_0}{9\cdot6\cdot3\cdot8}$
$a_{12}=-\dfrac8{12\cdot11}a_9=-\dfrac{a_0}{12\cdot9\cdot6\cdot3\cdot11}$

and so on. The reordering of the product in the denominator is intentionally done to make the pattern clearer. We can surmise the general pattern for $n=3k$ as

$a_{3k}=\dfrac{(-1)^{k+1}a_0}{(3k\cdot(3k-3)\cdot(3k-2)\cdot\cdots\cdot6\cdot3\cdot(3k-1)}$
$a_{3k}=\dfrac{(-1)^{k+1}a_0}{3^k(k\cdot(k-1)\cdot\cdots\cdot2\cdot1)\cdot(3k-1)}$
$a_{3k}=\dfrac{(-1)^{k+1}a_0}{3^kk!(3k-1)}$

So the series solution to the ODE is given by

$y=\displaystyle\sum_{n\ge0}a_nx^n$
$y=a_1x+\displaystyle\sum_{k\ge0}\frac{(-1)^{k+1}a_0}{3^kk!(3k-1)}$

Attached is a plot of a numerical solution (blue) to the ODE with initial conditions sampled at $a_0=y(0)=1$ and $a_1=y'(0)=2$ overlaid with the series solution (orange) with $n=3$ and $n=6$ . (Note the rapid convergence.)

7 0

1 year ago