Too bad you haven't shared illustrations of the possible solutions.
Starting with 4|x+3|>8, div. both sides by 4: <span>|x+3|>2
Case 1: x+3 is already positive: then x+3>2, or x > -1
Case 2: x+3 is negative: Then -(x+3)>2, or -x - 3 > 2, or -5>x or x<-5
Draw a number line representing x values. Place empty circles at x = -5 and x = -1. Draw a vector from the circle at x = -1, to the right of x = -1. Draw a vector from the circle at x = - 5, to the left of x = -5. Note that x values between x = -5 and x = -1 are not solutions.</span>
<span>1. Evaluate (2-2^3)^3-4x(-4)
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![(2-2 x^{3} )^{3} -4x(-4)=8-8x^{9} -16x](https://tex.z-dn.net/?f=%282-2%20x%5E%7B3%7D%20%29%5E%7B3%7D%20-4x%28-4%29%3D8-8x%5E%7B9%7D%20-16x)
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![=8(1-x^{9} -2x)](https://tex.z-dn.net/?f=%3D8%281-x%5E%7B9%7D%20-2x%29)
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2. Compute
-453+(-5.34)=-453-5.34
=-458.34
-56+27.4=-28.6
3. Use <, >, or =
9 > -8
-12 < -3
-4 > -7</span>
So first y do is is add 5+7 and add 3/10 + 7/10 equal 10/10 so 5+7 + 1 is 13 she used 13 bags of flour the second week she used 1 more than last week
Given:
![x=4y+3,y=4+t](https://tex.z-dn.net/?f=x%3D4y%2B3%2Cy%3D4%2Bt)
Required: Parametric equation of y.
Explanation:
Substitute 4+t for y into the equation of x.
![\begin{gathered} x=4(4+t)+3 \\ =16+4t+3 \\ =4t+19 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%3D4%284%2Bt%29%2B3%20%5C%5C%20%3D16%2B4t%2B3%20%5C%5C%20%3D4t%2B19%20%5Cend%7Bgathered%7D)
So, the parametric equation for x is x = 4t+19.
Final Answer: The parametric equation of x is x = 4t + 19.