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3 years ago

I NEED HELP ASAP THANKSSS factor x^2 +9x+20

Mathematics

2 answers:

matrenka [14]3 years ago

6 0

Answer:

(x + 5)(x + 4)

Step-by-step explanation:

$x^{2} +9x+20=(x+5)(x+4)$

Reika [66]3 years ago

6 0

Answer:

Step-by-step explanation:

x^2+9x+20

(x+4)(x+5)

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Ok so i need help with this the image is below

ELEN [110]

Answer: the line shown in the graph is different

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

"What is the probability that the next two employees that join the office staff will take the bus"

castortr0y [4]

Answer:

The correct option is A.

Step-by-step explanation:

Consider the complete question is "The table shows the transportation method used by the employees in an office.

Transportation Number of employees

Car 18

Bus 6

Walk 6

what is the probability that the next two employees that join the office staff will take the bus? A.0.04, B.0.12, C.0.33, D.0.36.

Total number of employees = 18 + 6 + 6 = 30

The probability that the employees take the bus = $\frac{6}{30}=0.2$

The probability that the employees that join the office staff will take the bus is

$Probability=(0.2)\times (0.2)=0.04$

Therefore, the correct option is A.

7 0

3 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 $y = e^x$

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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.