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3 years ago

Can someone help me with this math homework please!

Mathematics

1 answer:

Daniel [21]3 years ago

8 0

Answer:

1st, 2nd and 6th options

Step-by-step explanation:

Given

$\frac{2}{3}$ - x + $\frac{1}{6}$ = 6x ( multiply through by 6 to clear the fractions )

4 - 6x + 1 = 36x ← option 1 will have the same solution set

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Adding the 2 fractions on the left side gives

$\frac{4}{6}$ + $\frac{1}{6}$ - x = 6x , that is

$\frac{5}{6}$ - x = 6x ← option 2 will have the same solution set

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

From

4 - 6x + 1 = 36x ( add 6x to both sides )

5 = 42x ← option 6 will have the same solution set

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A two digit number when read from left to right to right consiststs of consecutive integers. the nmber is six more than for time

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I suppose you want to know such number. Since we have a two digit number consisting of two consecutive integers, the only possible numbers are:

$12,23,34,45,56,67,78,89$

Since we sorted all the cases out, we simply have to check which one satisfies the requirement. For each number, we'll write four times the the sum of its digits, and add 6, hoping to get the original number.

$\text{Number: } 12,\ \ 4\times \text{sum of digits: } 12\ \ \text{plus six: } 18\neq 12$

$\text{Number: } 23,\ \ 4\times \text{sum of digits: } 24\ \ \text{plus six: } 30\neq 23$

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3 years ago

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$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

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.

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3 years ago

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