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erik [133]
3 years ago
10

Write sentence as inequality. Plz help A number x is less than 5 and greater than 3.

Mathematics
1 answer:
Y_Kistochka [10]3 years ago
4 0

Answer:

3<x<5

Step-by-step explanation:

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How would you solve for the question below???<br> step by step please and thanks.
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Combine the two equations in the right amounts to eliminate y :

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Find the coordinates of the point three tenths<br><br> of the way from A(-4, -8) to B(11, 7)
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let's first off take a peek at those values.

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\bf ~~~~~~~~~~~~\textit{internal division of a line segment}&#10;\\\\\\&#10;A(-4,-8)\qquad B(11,7)\qquad&#10;\qquad \stackrel{\textit{ratio from A to B}}{3:7}&#10;\\\\\\&#10;\cfrac{A\underline{C}}{\underline{C} B} = \cfrac{3}{7}\implies \cfrac{A}{B} = \cfrac{3}{7}\implies 7A=3B\implies 7(-4,-8)=3(11,7)\\\\[-0.35em]&#10;~\dotfill\\\\&#10;C=\left(\frac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \frac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)\\\\[-0.35em]&#10;~\dotfill


\bf C=\left(\cfrac{(7\cdot -4)+(3\cdot 11)}{3+7}\quad ,\quad \cfrac{(7\cdot -8)+(3\cdot 7)}{3+7}\right)&#10;\\\\\\&#10;C=\left( \cfrac{-28+33}{10}~~,~~\cfrac{-56+21}{10} \right)\implies C=\left( \cfrac{5}{10}~~,~~\cfrac{-35}{10} \right)&#10;\\\\[-0.35em]&#10;\rule{34em}{0.25pt}\\\\&#10;~\hfill C=\left( \frac{1}{2}~,~-\frac{7}{2} \right)~\hfill

4 0
3 years ago
In problem solve the given differential equation by underdetermined coefficients y''-2y+y=xe^x
Alexus [3.1K]

Answer:

Solution is y(t)=C_1e^x+C_2xe^x+\frac{x^3e^x}{6}

Step-by-step explanation:

Given Differential Equation,

y"-2y'+y=xe^x ...............(1)

We need to solve the given differential equations using undetermined coefficients.

Let the solution of the given differential equation is made up of two parts. one complimentary solution and one is particular solution.

\implies\:y(x)=y_c(x)+y_p(x)

For Complimentary solution,

Auxiliary equation is as follows

m² - 2m + 1 = 0

( m - 1 )² = 0

m = 1 , 1

So,

y_c(x)=C_1e^x+c_2xe^x

Now for particular solution,

let y_p(x)=Ax^3e^x

y'=Ax^3e^x+3Ax^2e^x

y"=Ax^3e^x+6Ax^2e^x+6Axe^x

Now putting these values in (1), we get

Ax^3e^x+6Ae^2e^x+6Axe^x-2(Ax^3e^x+3Ax^2e^x)+Ax^3e^x=xe^x

Ax^3e^x+6Ae^2e^x+6Axe^x-2Ax^3e^x-6Ax^2e^x+Ax^3e^x=xe^x

6Axe^x=xe^x

6A=1

A=\frac{1}{6}

\implies\:y_p(x)=\frac{x^3e^x}{6}

Therefore, Solution is y(t)=C_1e^x+C_2xe^x+\frac{x^3e^x}{6}

8 0
4 years ago
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