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2 years ago

Help plz..don’t steal points

Mathematics

2 answers:

Yuliya22 [10]2 years ago

4 0

It’s a whole number. Should be if following standard rules. In decimals anything before the dot is a while number so if it’s 2.75 it’s 2 cookies and 75 percent of cookies if it was 3.00 then it’s three cookies

alexdok [17]2 years ago

3 0

I think it’s Integer I’m not sure

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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

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

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

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2 years ago

The cost of six hens and one duck at the university poultry farm is GH₵40. While four hens and three ducks cost GH₵36. What is t

Anon25 [30]

Answer:

Step-by-step explanation:

Let's call hens h and ducks d. The first algebraic equation says that 6 hens (6h) plus (+) 1 duck (1d) cost (=) 40.

The second algebraic equations says that 4 hens (4h) plus (+) 3 ducks (3d) cost (=) 36.

The system is

6h + 1d = 40

4h + 3d = 36

The best way to go about this is to solve it by substitution since we have a 1d in the first equation. We will solve that equation for d since that makes the most sense algebraically. Doing that,

1d = 40 - 6h.

Now that we know what d equals, we can sub it into the second equation where we see a d. In order,

4h + 3d = 36 becomes

4h + 3(40 - 6h) = 36 and then simplify. By substituting into the second equation we eliminated one of the variables. You can only have 1 unknown in a single equation, and now we do!

4h + 120 - 18h = 36 and

-14h = -84 so

h = 6.

That means that each hen costs $6. Since the cost of a duck is found in the bold print equation above, we will sub in a 6 for h to solve for d:

1d = 40 - 6(6) and

d = 40 - 36 so

d = 4.

That means that each duck costs $4.

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