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3 years ago

HELPP -1-3(5m+8) ≥-85 i need help :D

Mathematics

1 answer:

forsale [732]3 years ago

6 0

Answer:

- 1 - 3(5m + 8) ≥ -85

-1 - 15m - 24 ≥ -85

-15m ≥ -85 + 1 + 24

-15m ≥ 25 - 85

-15m ≥ -60

(-1)(-15)m ≤ -60(-1)

15m ≤ 60

m ≤ 4

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Dan was getting rs. 100 for 5 years at an interest rate of 5%. Find the accumulated value of annuity at the end of 5 years

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2 years ago

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3 years ago

Read 2 more answers

Consider the differential equation

Ainat [17]

Answer:

$W\left ( e^{\frac{x}{2}},xe^{\frac{x}{2}} \right )=e^x$

Step-by-step explanation:

Let $y=e^{\frac{x}{2}}$

Differentiate with respect to $x$

$y'=\frac{1}{2}e^{\frac{x}{2}}$

Differentiate with respect to $x$

$y''=\frac{1}{4}e^{\frac{x}{2}}$

Put values of $y,y',y''$ in $4y''-4y'+y=0$

$4y''-4y'+y=0\\4\left (\frac{1}{4}e^{\frac{x}{2}} \right )-4\left ( \frac{1}{2}e^{\frac{x}{2}}\right )+e^{\frac{x}{2}}\\=e^{\frac{x}{2}}-2e^{\frac{x}{2}}+e^{\frac{x}{2}}\\=2e^{\frac{x}{2}}-2e^{\frac{x}{2}}\\=0$

So, $y=e^{\frac{x}{2}}$ is the solution of the given equation.

Now, let $y=xe^{\frac{x}{2}}$

Differentiate with respect to $x$

$y'=e^{\frac{x}{2}}+\frac{x}{2}e^{\frac{x}{2}}=e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right )$

Differentiate with respect to $x$

$y''=\frac{1}{2}e^{\frac{x}{2}}+\frac{1}{2}e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right )=e^{\frac{x}{2}}+\frac{1}{4}xe^{\frac{x}{2}}$

Put values of $y,y',y''$ in $4y''-4y'+y=0$

$4y''-4y'+y=0\\4\left (e^{\frac{x}{2}}+\frac{1}{4}xe^{\frac{x}{2}} \right )-4\left ( e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right )\right )+xe^{\frac{x}{2}}\\=4e^{\frac{x}{2}}+xe^{\frac{x}{2}}-2e^{\frac{x}{2}}(2+x)+xe^{\frac{x}{2}}\\=4e^{\frac{x}{2}}+xe^{\frac{x}{2}}-4e^{\frac{x}{2}}-2xe^{\frac{x}{2}}+xe^{\frac{x}{2}}\\=0$

To find: $W\left ( e^{\frac{x}{2}},xe^{\frac{x}{2}} \right )$

Solution:

Let $u=e^{\frac{x}{2}}\,,\,v=xe^{\frac{x}{2}}$

$W(u,v)=\left | \begin{matrix}u&v\\u'&v' \end{matrix} \right |\\=\left | \begin{matrix}e^{\frac{x}{2}}&xe^{\frac{x}{2}}\\\frac{1}{2}e^{\frac{x}{2}}&e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right ) \end{matrix} \right |\\=e^{\frac{x}{2}}\left [ e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right ) \right ]-\frac{1}{2}e^{\frac{x}{2}}xe^{\frac{x}{2}}\\=e^x\left ( 1+\frac{x}{2} \right )-\frac{1}{2}xe^x\\=e^x+\frac{1}{2}xe^x-\frac{1}{2}xe^x\\=e^x$

5 0

3 years ago