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4 years ago

Find the exterior angle of a 15-gon

1 answer:

3 0

$\bf \textit{sum of all exterior angles in a polygon}\\\\ n\theta =360~~ \begin{cases} n=\textit{number of}\\ \qquad sides\\ \theta =angle~in\\ \qquad degrees\\[-0.5em] \hrulefill\\ n=15 \end{cases}\implies 15\theta =360\implies \theta =\cfrac{360}{15}\implies n=24$

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Multiply the following and write your answer In scientific notation.. 3.62 × 10^-5 · 9,400.

svet-max [94.6K]

Answer:

3.4028 x 10^-1

Step-by-step explanation:

First you make the 9400 into scientific notation:

9.4x10^3

Then you combine like terms

(3.62x9.4)(10^-5x10^3)

Finally you simplify

34.028x10^-2

3.4028x10^-1

6 0

3 years ago

A movie collector has 105 movie posters and 90 band posters if she sells 21 movie posters then how many band posters should she

kotykmax [81]

Hi there! :)

Answer:

In order to maintain the same ratio, she should sell 18 band posters.

Step-by-step explanation:

Initial ratio:

Ratio is: 105 movie posters for 90 band posters = 105 : 90

She sells 21 movie posters, which means that we are left with 105 movies minus the 21 she sold.

105 - 21 = 84 → She now has 84 movie posters.

In order to answer your question, you'll need to use the cross product method:

105 movie posters → 90 band posters

84 movie posters → X

(90 × 84) ÷ 105 = X

7,560 ÷ 105 = X

72 = X ⇒ Number of band posters she should have left to maintain the same ratio.

She has 90 band posters :

90 - X = 72

Add "X" to each side of the equation → 72 + X = 72 + X

90 = 72 + X

Subtract 72 from each side of the equation → 90 - 72 = 18

18 = X

There you go! I really hope this helped, if there's anything just let me know! :)

4 0

4 years ago

What's the place value of 7 in 3,729,760

Vadim26 [7]

The value of the 7 are

one of them is one hundred thousand

the other one is one hundred

3 0

3 years ago

Read 2 more answers

Let y(t) be the solution to y˙=3te−y satisfying y(0)=3 . (a) Use Euler's Method with time step h=0.2 to approximate y(0.2),y(0.4

OLEGan [10]

Answer:

$y(0.2)=3$ , $y(0.4)=3.005974448$ , $y(0.6)=3.017852169$ , $y(0.8)=3.035458382$ , and $y(1.0)=3.058523645$

The general solution is $y=\ln \left(\frac{3t^2}{2}+e^3\right)$

The error in the approximations to y(0.2), y(0.6), and y(1):

$|y(0.2)-y_{1}|=0.002982771$

$|y(0.6)-y_{3}|=0.008677796$

$|y(1)-y_{5}|=0.013499859$

Step-by-step explanation:

Point a:

The Euler's method states that:

$y_{n+1}=y_n+h \cdot f \left(t_n, y_n \right)$ where $t_{n+1}=t_n + h$

We have that $h=0.2$ , $t_{0}=0$ , $y_{0} =3$ , $f(t,y)=3te^{-y}$

We need to find $y(0.2)$ for $y'=3te^{-y}$ , when $y(0)=3$ , $h=0.2$ using the Euler's method.

So you need to:

$t_{1}=t_{0}+h=0+0.2=0.2$

$y\left(t_{1}\right)=y\left(0.2)=y_{1}=y_{0}+h \cdot f \left(t_{0}, y_{0} \right)=3+h \cdot f \left(0, 3 \right)=$

$=3 + 0.2 \cdot \left(0 \right)= 3$

$y(0.2)=3$

We need to find $y(0.4)$ for $y'=3te^{-y}$ , when $y(0)=3$ , $h=0.2$ using the Euler's method.

So you need to:

$t_{2}=t_{1}+h=0.2+0.2=0.4$

$y\left(t_{2}\right)=y\left(0.4)=y_{2}=y_{1}+h \cdot f \left(t_{1}, y_{1} \right)=3+h \cdot f \left(0.2, 3 \right)=$

$=3 + 0.2 \cdot \left(0.02987224102)= 3.005974448$

$y(0.4)=3.005974448$

The Euler's Method is detailed in the following table.

Point b:

To find the general solution of $y'=3te^{-y}$ you need to:

Rewrite in the form of a first order separable ODE:

$e^yy'\:=3t\\e^y\cdot \frac{dy}{dt} =3t\\e^y \:dy\:=3t\:dt$

Integrate each side:

$\int \:e^ydy=e^y+C$

$\int \:3t\:dt=\frac{3t^2}{2}+C$

$e^y+C=\frac{3t^2}{2}+C\\e^y=\frac{3t^2}{2}+C_{1}$

We know the initial condition y(0) = 3, we are going to use it to find the value of $C_{1}$

$e^3=\frac{3\left(0\right)^2}{2}+C_1\\C_1=e^3$

So we have:

$e^y=\frac{3t^2}{2}+e^3$

Solving for y we get:

$\ln \left(e^y\right)=\ln \left(\frac{3t^2}{2}+e^3\right)\\y\ln \left(e\right)=\ln \left(\frac{3t^2}{2}+e^3\right)\\y=\ln \left(\frac{3t^2}{2}+e^3\right)$