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3 years ago

Solve the equation: 5|2x-7|=20

Mathematics

2 answers:

MA_775_DIABLO [31]3 years ago

6 0

Answer:

x = $\frac{11}{2}$ , $\frac{3}{2}$

Step-by-step explanation:

First, isolate the absolute value:

5|2x - 7| = 20

divide both sides by 5:

|2x - 7| = 4

break into 2 equations:

2x - 7 = 4, 2x - 7 = -4

solve both equations and collect solutions:

2x = 11, 2x = 3

x = $\frac{11}{2}$ , $\frac{3}{2}$

Pepsi [2]3 years ago

6 0

Answer:

x=11/2,3/2

Step-by-step explanation:

Given equation:

5|2x-7|=20

=> |2x-7| = 20/5 = 4

Since the left hand side is an absolute value, the equation can be expressed in two alternate forms.

The two possible scenarios are:

1) 2x-7 = 4 or 2) 2x-7 = -4

Solving these two equations individually:

Equation 1:

1) 2x-7 = 4

=> 2x = 4+7=11

=> x= 11/2

2) 2x-7 = -4

=> 2x = 7-4 = 3

=> x=3/2

Hence the required values of x are 11/2,3/2

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Igoryamba

Answer:

The curvature is $\kappa=1$

The tangential component of acceleration is $a_{\boldsymbol{T}}=0$

The normal component of acceleration is $a_{\boldsymbol{N}}=1 (2)^2=4$

Step-by-step explanation:

To find the curvature of the path we are going to use this formula:

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$\boldsymbol{r}'(t)=\frac{d}{dt}\left(cos\left(2t\right)\right)\:\boldsymbol{i}+\frac{d}{dt}\left(sin\left(2t\right)\right)\:\boldsymbol{j}+\frac{d}{dt}\left(1)\right\:\boldsymbol{k}\\\boldsymbol{r}'(t)=-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}$

Next , we find the magnitude of derivative of the position vector

$|| \boldsymbol{r}'(t)}||=\sqrt{(-2\sin \left(2t\right))^2+(2\cos \left(2t\right))^2} \\|| \boldsymbol{r}'(t)}||=\sqrt{2^2\sin ^2\left(2t\right)+2^2\cos ^2\left(2t\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4\left(\sin ^2\left(2t\right)+\cos ^2\left(2t\right)\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4}\sqrt{\sin ^2\left(2t\right)+\cos ^2\left(2t\right)}\\\\\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)+\sin ^2\left(x\right)=1\\\\|| \boldsymbol{r}'(t)}||=2\sqrt{1}=2$

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$\boldsymbol{T}}=\frac{\boldsymbol{r}'(t)}{||\boldsymbol{r}'(t)||}$

$\boldsymbol{T}}=\frac{-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}}{2} =\sin \left(2t\right)+\cos \left(2t\right)$

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$a_{\boldsymbol{N}}=\kappa (\frac{ds}{dt})^2$

We know that $\kappa=1$ and $\frac{ds}{dt}=2$ so

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Yakvenalex [24]

Answer:

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Step-by-step explanation:

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