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Anuta_ua [19.1K]
3 years ago
9

Can someone help me solve for x in this equation?? 25+3.50x < 17+7.50x

Mathematics
1 answer:
Lelu [443]3 years ago
4 0
So we want to solve like if it was just a basic equation. We start with 25+3.50x<17+7.50x and get rid of our smallest x. That means we get rid of the 3.50x by subtracting from both sides and our new equation is 25<17+4x. Let’s get our x by itself shall we, so let’s subtract 17 from both sides and we get 8<4x. Finally all we do is divide by 4 and just like that 2
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Find parametric equations and symmetric equations for the line. (Use the parameter t.) The line through (4, −5, 2) and parallel
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Answer:

Step-by-step explanation:

From the given information, the symmetric equations for the line pass through(4, -5, 2) i.e (x_o, y_o, z_o) and are parallel to \dfrac{x+5}{1} = \dfrac{y}{2}= \dfrac{z-3}{1}

The parallel vector to the line i + zj+k = ai + bj + ck

Hence, the equation for the line is :

x = x_o + at \\ \\ x = y_o + bt \\ \\ x = z_o + ct

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y = -5 + 2t

z = 2 + t

Thus, x, y, z = ( 4+t, -5+2t, 2+t )

The symmetric equation can now be as follows:

\begin  {vmatrix} x = 4+ t   \\ \\  \dfrac{x-4}{1} = t  \begin {vmatirx} \end {vmatrix}\begin {vmatrix} y = - 5+2t  \\ \\ \dfrac{y+5}{2}  =t      \end {vmatrix}\begin {vmatrix} z =2+t  \\ \\ \dfrac{z-2}{1}  =t      \end {vmatrix}

∴

\dfrac{x-4}{1}= \dfrac{y+5}{2}=\dfrac{z-2}{1}

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Exact Form:

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Decimal Form:

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Step-by-step explanation:

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