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3 years ago

Please Help , Thank You

Mathematics

2 answers:

olya-2409 [2.1K]3 years ago

6 0

I'll go with A and B

Hope this helps

Ad libitum [116K]3 years ago

4 0

The second one is the answer because the equations are exactly the same

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Firdavs [7]

Basically, you just want to keep flipping a coin a total of 10 times, 27 more times. (If that makes any sense) You also want to keep recording the number of heads in each set of 10.

3 0

3 years ago

The picture is attached with t

weeeeeb [17]

Your answers should be B and D

When naming angles, they all need to be connected on the line without overlapping any points previously said. Angle YXZ is incorrect because X is not on the same line as Z. YW is a line because both sides continue on forever, as indicated by the arrows.

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6 0

2 years ago

It has been said that newspapers are disappearing, replaced by various electronic media. In 2004, newspaper circulation (the num

shutvik [7]

Answer:

A. $y=-1.2x+59$

B. 42.2 million

Step-by-step explanation:

Part A:

Given:

Newspapers circulated in year 2004, $y_{1}=59\textrm{ million}$

Newspapers circulated in year 2014, $y_{2}=47\textrm{ million}$

Let the time $x$ start at the year 2004. So, $x_{1}=0$

For the year 2014, $x_{2}=2014-2004=10$

Therefore, linear relationship between newspapers circulated and time passed since 2004 is given as:

$y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})\\y-59=\frac{47-59}{10-0}(x-0)\\y-59=-\frac{12}{10}x\\y=-1.2x+59$

Therefore, the equation describing the relationship is: $y=-1.2x+59$

Part B:

For the year 2018, $x=2018-2004=14$

Plug in 14 for $x$ in the above equation and solve for $y$ . This gives,

$y=-1.2(14)+59\\y=-16.8+59\\y=42.2$

Therefore, in the year 2018, the newspaper circulation will be 42.2 million.

8 0

2 years ago

Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

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 $y = e^x$

-----------------------

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

Can someone help on my math test? giving out brainiest!

katovenus [111]

Answer:

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8 0

3 years ago

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