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3 years ago

10

Hey can somebody help me with this please and thank yu. :)

1 answer:

Arturiano [62]3 years ago

3 0

Answer:

x= -5

hope it's helpful ❤❤❤

THANK YOU

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0.345 recurring as a fraction please - the 3 and 5 are recurring

Aneli [31]

Answer:

The decimal 0.345 expressed as a fraction is 69200 .

8 0

2 years ago

The answer to this problem

4vir4ik [10]

$2 \frac{6}8]$

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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

What is the expression obtained when the sum of 2a−16abc−45 and 2a+7abc+11 is subtracted from 6a+6abc−30?

kow [346]

Answer:

$2a+15abc+4$

Step-by-step explanation:

$(2a-16abc-45)+(2a+7abc+11)$

$\implies 2a-16abc-45+2a+7abc+11$

collect and combine like terms:

$\implies 4a-9abc-34$

$(6a+6abc-30)-(4a-9abc-34)$

Apply the distributive law $-\left(a-b\right)=-a+b$ :

$\implies 6a+6abc-30-4a+9abc+34$

collect and combine like terms:

$\implies 2a+15abc+4$

7 0

2 years ago

What are the numbers of symmetries for this figure? Enter your answers in the boxes. Number of rotational symmetries: Number of

Ann [662]

Answer:

??? and 0

Step-by-step explanation:

Ok I don't know about the first one but all I know is that it has 0 reflectional symmetries. Apologies for that but I wish you luck on your work and hopefully this somewhat helps a bit. ;-D

3 0

3 years ago