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3 years ago

12

a road sign is in the shape of a regular pentagon what is the measure of each angel on the sign round to the nearest tenth

1 answer:

yaroslaw [1]3 years ago

3 0

Answer:

Each angle is equal to 108 degrees. The internal angles of a pentagon add up to 540.

No matter what the side lengths are, so long as it is a regular pentagon, each angle will always be exactly 108 degrees.

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What is the best measure of center for the following data set?

Rom4ik [11]

Answer:

b) median: there are 7 numbers listed; 20 is the middle number among them

range: subtract lowest from highest number 25-2=23

mean: add all the numbers together =129

mode: place numbers in order (smallest to greatest), then see which number appears the most 2, 18, 18, 20, 22, 24, 25=18

Step-by-step explanation:

5 0

2 years ago

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in <em>c)</em>

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution <em>m=α±βi</em> where <em>α=0</em> and <em>β=2</em>

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where <em>A</em> and <em>B</em> are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

7 0

3 years ago

If m angle BGF=152 degrees, what is m angle AGF

Allushta [10]

The Answer is: 64 degrees

4 0

3 years ago

Read 2 more answers

Does anyone know how to answer these two questions?

Lera25 [3.4K]

2c+5=c+8
2c-c=8-5
1c=3
c=3/1
c=3

3 0

3 years ago

Given that <img src="https://tex.z-dn.net/?f=2%5Ex" id="TexFormula1" title="2^x" alt="2^x" align="absmiddle" class="latex-formul

sweet-ann [11.9K]

Answer:

$\dfrac{49}{4} = 12.25$

Step-by-step explanation:

$4^{x-1} =$

$= (2^2)^{x - 1}$

$= 2^{2(x - 1)}$

$= 2^{2x - 2}$

$= 2^{x + x - 2}$

$= 2^x \times 2^x \times 2^{-2}$

$= 7 \times 7 \times \dfrac{1}{2^2}$

$= \dfrac{49}{4} = 12.25$

7 0

2 years ago