Ask question

Ask question

All categories

Salsk061 [2.6K]

2 years ago

8

Camille knew that a triangle had one side with a length of 16 inches and another side with a length of 20 inches. She did not kn

ow the length of the third side, but she did know that the perimeter was five times the length of the unknown side. How long is the unknown side? Please I really need help with this question!!

1 answer:

SSSSS [86.1K]2 years ago

5 0

The unknown side is 9. Because 9 x 5= 45, 45 being the perimeter, that is 5 times the unknown side. And 45 is the sum of 9, 16, and 20.

You might be interested in

The height of a cone-shaped container is 15 centimeters and its radius is 14 centimeters. Kate fills the container completely wi

lesantik [10]

The volume of the cone with radius r=14 cm and height h=15 cm is,
$V= \frac{1}{3} \pi r^2h = \frac{1}{3} \pi *14^2*15 = 3077.2~cm^3$

Each day 40 cm^3 is subtracted from the volume. So the volume of honey left after [d] number of days would be the starting volume minus 40 times number of days passed.
$V left=V start-40*d$
$V left = 3077.2 - 40d$

It asks when the volume will be empty, The volume left is zero after how many days?
$0 = 3077.2 - 40d$
d=76.93 days, or 80 days rounding to whole numbers.

7 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

2 years ago

Three trigonometric functions for a given angle are shown below. cosecant theta = thirteen-twelfths; secant theta = Negative thi

Kay [80]

Answer:

(–5, 12) is the correct answer.

Step-by-step explanation:

We are given the following values:

$cosec\theta = \dfrac{13}{12}\\sec\theta=-\dfrac{13}{5}\\cot\theta =-\dfrac{5}{12}$

Now, we know the following identities:

$sin \theta = \dfrac{1}{cosec\theta}\\cos \theta = \dfrac{1}{sec\theta}\\tan \theta = \dfrac{1}{cot\theta}$

Now, the values are:

$sin\theta = \dfrac{12}{13}\\cos\theta=-\dfrac{5}{13}\\tan\theta =-\dfrac{12}{5}$

Sine value is positive and cos, tan values are negative.

It can be clearly observed that $\theta$ is in 2nd quadrant.

2nd quadrant means, the value of $x$ will be negative and $y$ will be positive.

Let us have a look at the value of $tan\theta$ :

$tan\theta = \dfrac{Perpendicular}{Base}\\OR\\tan\theta = \dfrac{y-coordinate}{x-coordinate} = -\dfrac{12}{5}\\\therefore y = 12,\\x = -5$

Please refer to the attached image for clear understanding and detailed explanation.

Hence, the correct answer is coordinate (x,y) is (–5, 12)

8 0

1 year ago

The table below shows the total number of hours, Jessica, Emily, and Trina, each studied after given number of weeks. Part A one

pishuonlain [190]

Answer:Y times X = 5

Step-by-step explanation:

8 0

1 year ago

Read 2 more answers

Jimmy’s Delicatessen sells large tins of Tom Tucker’s Toffee. The deli uses a periodic review system, checking inventory levels

Yakvenalex [24]

Answer:

The restocking level is 113 tins.

Step-by-step explanation:

Let the random variable <em>X</em> represents the restocking level.

The average demand during the reorder period and order lead time (13 days) is, <em>μ</em> = 91 tins.

The standard deviation of demand during this same 13- day period is, <em>σ</em> = 17 tins.

The service level that is desired is, 90%.

Compute the <em>z</em>-value for 90% desired service level as follows:

$z_{\alpha}=z_{0.10}=1.282$

*Use a <em>z</em>-table for the value.

The expression representing the restocking level is:

$X=\mu +z \sigma$

Compute the restocking level for a 90% desired service level as follows:

$X=\mu +z \sigma$

$=91+(1.282\times 17)\\=91+21.794\\=112.794\\\approx 113$

Thus, the restocking level is 113 tins.

6 0

2 years ago