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3 years ago

Factor: a² + 9a+20 A (a-4)(-5) B (a+2)(a +10) C (-2)(a-10) D (a+4)(a +5)

Mathematics

2 answers:

konstantin123 [22]3 years ago

8 0

Answer: is D

Answer is D

Nina [5.8K]3 years ago

6 0

Using the simple method of quadratic expression:

we find two numbers(the first and last word) that when we multiply is the same as when we add them together.

a²+9a+20

a²+4a+5a+20

The answer is D

a(a+4)+5(a+4)

(a+4)(a+5)

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PICTURE ABOVE !!! <br> find the length of XZ <br> PLEASE HELP !!

slamgirl [31]

Answer: 60

Step-by-step explanation: 5 was multiplied by 6 to get 30 so that means 10 has to get multiplied by 6 and it turns into 60

4 0

3 years ago

Differentiate x^2 e^x logx

zimovet [89]

Product rule:

$\dfrac{\mathrm d}{\mathrm dx}(x^2e^x\log x)$

$=\dfrac{\mathrm d(x^2)}{\mathrm dx}e^x\log x+x^2\dfrac{\mathrm d(e^x)}{\mathrm dx}\log x+x^2e^x\dfrac{\mathrm d(\log x)}{\mathrm dx}$

Power rule:

$\dfrac{\mathrm d(x^2)}{\mathrm dx}=2x$

The exponential function is its own derivative:

$\dfrac{\mathrm d(e^x)}{\mathrm dx}=e^x$

Assuming the base of $\log x$ is $e$ , its derivative is

$\dfrac{\mathrm d(\log x)}{\mathrm dx}=\dfrac1x$

But if you mean a logarithm of arbitrary base $b$ , we have

$y=\log_bx\implies x=b^y=e^{y\ln b}\implies1=e^{y\ln b}\ln b\dfrac{\mathrm dy}{\mathrm dx}$

$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{e^{-y\ln b}}{\ln b}=\dfrac1{b^y\ln b}$

$\implies\dfrac{\mathrm d(\log_bx)}{\mathrm dx}=\dfrac1{x\ln b}$

So we end up with

$2xe^x\log x+x^2e^x\log x+\dfrac{x^2e^x}x$

$=xe^x(2\log x+x\log x+1)$

8 0

3 years ago

Identify the expression with nonnegative limit values. More info on the pic. PLEASE HELP.

marshall27 [118]

Answer:

$\lim _{x\to 2}\:\frac{x-2}{x^2-2}\\\\ \lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}\\\\ \lim _{x\to \frac{5}{2}}\left\frac{2x^2+x-15}{2x-5}\right$

Step-by-step explanation:

a) $\lim _{x\to 3}\:\frac{x^2-10x+21}{x^2+4x-21}=\lim \:_{x\to \:3}\:\frac{\left(x-7\right)\left(x-3\right)}{\left(x+7\right)\left(x-3\right)}=\lim \:_{x\to \:3}\:\frac{x-7}{x+7}=\frac{3-7}{3+7}=-\frac{4}{10}=-\frac{2}{5}$

b) $\lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=\lim \:_{x\to -\frac{3}{2}}\:\frac{\left(2x+3\right)\left(x-4\right)}{\left(2x+3\right)}=\lim \:\:_{x\to \:-\frac{3}{2}}\:\left(x-4\right)=-\frac{3}{2}-4\\ \\ \lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=-\frac{11}{2}$

c) $\lim _{x\to 2}\:\frac{x-2}{x^2-2}=\frac{2-2}{\left(2\right)^2-2}=\frac{0}{4-2}=0$

d) $\lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}=\lim _{x\to 11}\:\frac{\left(x-11\right)\left(x+17\right)}{\left(x-11\right)\left(x+14\right)}=\lim _{x\to 11}\:\frac{\left(x+17\right)}{\left(x+14\right)}=\frac{11+17}{11+14}=\frac{28}{25}$

e) $\lim _{x\to 3}\:\frac{x^2-8x+15}{x-3}=\lim \:_{x\to \:3}\:\frac{\left(x-3\right)\left(x-5\right)}{x-3}=\lim _{x\to 3}\left(x-5\right)=3-5=-2$

f) $\lim _{x\to \frac{5}{2}}\left(\frac{2x^2+x-15}{2x-5}\right)=\lim \:_{x\to \:\frac{5}{2}}\frac{\left(2x-5\right)\left(x+3\right)}{2x-5}=\lim \:\:_{x\to \:\:\frac{5}{2}}\left(x+3\right)=\frac{5}{2}+3=\frac{11}{2}$