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2 years ago

Please help i need to find the constant of proportionality

Mathematics

1 answer:

abruzzese [7]2 years ago

3 0

Answer:

Step-by-step explanation:

y = 4*x

y = 4*1.5 = 6

y = 4*2.25 = 9

y = 4* 3.25 = 13

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SVETLANKA909090 [29]

You should try C I’m not sure but try I guess!

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2 years ago

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Alex_Xolod [135]

Answer:B

Step-by-step explanation:

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3 years ago

in the year 2010, the luxury bike industry had two bike manufactures splendor and passion with the market shares of 30% and 70%,

kari74 [83]

Answer: 50%

Step-by-step explanation:

In 2010, there were only 2 bike manufacturers and they had market shares of 30% and 70%.

A new manufacturer joins the market and captures 10% of the market share and Splendor increases to 40% of market share.

Market shares:

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= 50%

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3 years ago

What is 7/8 divided by 11/16

Igoryamba

1.27 repeated is the answer

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3 years ago

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SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

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Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

6 0

3 years ago