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mylen [45]
3 years ago
9

Which ratio is equal to 5/6 and Which ratio is not equal to 4/9? ...?

Mathematics
1 answer:
Tatiana [17]3 years ago
8 0
It seems that you have missed the relevant options for your question, but hope these answers can help you find the one. Anyway, the ratio that can be equal to 5/6 is 10/12. Or it can be the following: 15/18, 20/24 or 25/30.
The ratio that is not equal to 4/9 among the given choices which are <span>45/95,   16/36,   8/18,   20/45 is 45/95. The answer is the first option. Since 45/95 is 9/19. Hope these answers help.</span>
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A population of 50 timber wolves was introduced into a forest in 2009. The population is expected to grow by the function p(t)=5
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The population is expected to grow by the function p(t)=50e.85t, where t is in years.a) What will be the population in 2015?

Step-by-step explanation:

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You and 3 friends are going to Great Mountain Park. The park has 3 water rides. The Coaster, Slide, and Drop cost $5.50, $2.00,
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Alana raked leaves for 7 days in the fall. She raked 9 bags of leaves each day. Alan earned d dollars for each bag she raked. Wh
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4 years ago
an ellipse has a center at the origin, a vertex along the major axis at (13, 0), and a focus at (12, 0). What is the equation of
yan [13]
Hi, as the elipce has a center at the origion and tha major distance is the axis

\frac{x^2}{a^2} + \frac{y^2}{b^2}  = 1

But, 

c = 12        Distance of focus

a = 13   ->  Major distance

b = ?  -> 

But, 

\\ b^2 + c^2 = a^2&#10; \\ &#10; \\ b^2 + 12^2 = 13^2&#10; \\ &#10; \\ b^2 + 144 = 169&#10; \\ &#10; \\ b^2 = 169 - 144&#10; \\ &#10; \\ b^2 = 25&#10; \\ &#10; \\ b =  \sqrt{5} &#10; \\ &#10; \\ b = 5&#10;&#10;

Then,


\\  \frac{x^2}{(13)^2} +  \frac{y^2}{(5)^2} = 1&#10; \\ &#10; \\ or&#10; \\ &#10; \\  \frac{x^2}{169} +  \frac{y^2}{25} = 1
5 0
3 years ago
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Find the general solution of the following equation: y'(t) = 3y -5
Anton [14]

Answer:

The general solution of the equation is y = \frac{A}{3}e^{3t} + 5

Step-by-step explanation:

Since the differential equation is given as y'(t) = 3y -5

The differential equation is re-written as

dy/dt = 3y - 5

separating the variables, we have

dy/(3y - 5) = dt

dy/(3y - 5) = dt

integrating both sides, we have

∫dy/(3y - 5) = ∫dt

∫3dy/[3(3y - 5)] = ∫dt

(1/3)∫3dy/(3y - 5) = ∫dt

(1/3)㏑(3y - 5) = t + C

㏑(3y - 5) = 3t + 3C

taking exponents of both sides, we have

exp[㏑(3y - 5)] = exp(3t + 3C)

3y - 5 = e^{3t}e^{3C}        

3y - 5 = Ae^{3t}                A = e^{3C}

3y = Ae^{3t} + 5    

dividing through by 3, we have

y = \frac{A}{3}e^{3t} + 5

So, the general solution of the equation is y = \frac{A}{3}e^{3t} + 5

3 0
4 years ago
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