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Stells [14]
3 years ago
13

A map has a scale of 1 cm : 9 km. If two cities are 10 cm apart on the map, what is the actual distance between the cities, to t

he nearest tenth of a kilometer?
Mathematics
1 answer:
Ymorist [56]3 years ago
8 0
We can solve this problem by the rule of three:
1 cm-----------------9 Km
10 cm----------------  x

x=(10 cm * 9 km) / 1 cm=90.0 Km.

answer: 90.0 Km
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Y''+y'+y=0, y(0)=1, y'(0)=0
mars1129 [50]

Answer:

y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form ay''+by'+cy=0.

Given: y''+y'+y=0

Let y=e^{rt} be it's solution.

We get,

\left ( r^2+r+1 \right )e^{rt}=0

Since e^{rt}\neq 0, r^2+r+1=0

{ we know that for equation ax^2+bx+c=0, roots are of form x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} }

We get,

y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}

For two complex roots r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta, the general solution is of form y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )

i.e y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

Applying conditions y(0)=1 on e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right ), c_1=1

So, equation becomes y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

On differentiating with respect to t, we get

y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )

Applying condition: y'(0)=0, we get 0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}

Therefore,

y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

3 0
3 years ago
HELP ME WITH 16 & 17
damaskus [11]

__Data__

The equilateral's lengths are all the same

Also if AB = AD = BD = CD that would mean all those sides have the exact same length

Answers and Explanations

15A. BCD has 2 acute angles and one obtuse angle, so this is an Obtuse Triangle

15B. If beforementioned = 5 cm

Then the perimeter is 5cm x 3 = 15cm

15C. ABC is a Right Triangle  

16. abcdef. Measure the angles using a protractor

17. The shortest side of ABC is AB which is 5cm, the longest is AC which is 10cm

Shortest side: Longest side

5cm:10cm

1:2 is the ratio

7 0
2 years ago
Circumference of the cylinder a cylinder of the base with the diameter of 6cm and height of 12cm
yuradex [85]

Answer:

To find the volume, we must find the area of the circle and multiply it by its height, think of a cylinder as an object made up of MANY circles. So, V=pir^2h... so diameter is 6cm so radius = 3cm...

therefore, Volume = (pi)(3^2)(12) = 108(pi)

where 12 = height 3 = radius

For the surface area, we need to find the area of the 2 circles(top and bottom ) so once again,

2(pi)(r^2) = 2(pi)(9)

for the area of the rectangle(if we unfold the cylindrical can) we get D(pi)(12) where D= Diameter

so the surface area is

SA = (2)(9)(pi) + 6(pi)(12)

SA = 18pi + 72pi = 90pi

7 0
2 years ago
Read 2 more answers
Can Someone Explain??
ValentinkaMS [17]
A) Dependent Variable = Sue's Earning
Independent Variable = Number of hours she work

Y = 6x

B) When, x = 1, y = 6(1) = 6, 
when x = 2, y = 6(2) = 12
when x = 3, y = 6(3) = 18

Ordered pairs would be: (1, 6), (2, 12), (3, 18)

C) Equation would be: Y = 6x
Where, y = Earning
x = number of hours

Hope this helps!
3 0
3 years ago
Y = x^2 - 4 <br> y= x + 2
Snezhnost [94]
Hopes this help:

Answer: x = 3, -2 & y = 5, 0

Hopes this is the answer that you where looking for if not I am so sorry.
5 0
2 years ago
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