multiple choice 2 points\nwhat is the value of w in the equation \\( \\frac { 3 } { 4 } w + 8 = w - 7…

Question 28

Explanation:

Step1: Solve the equation

Given (\frac{3}{4}w + 8=w - 7). First, move the terms with (w) to one side: (8 + 7=w-\frac{3}{4}w).

Step2: Simplify the right - hand side

(15=\frac{4w-3w}{4}), since (w=\frac{4w}{4}). So (15=\frac{w}{4}).

Step3: Solve for (w)

Multiply both sides by (4): (w = 60).

Answer:

(60)

Question 29

Explanation:

Step1: Find the derivative of (y=(x^{2}-1)^{4})

Let (u = x^{2}-1), then (y = u^{4}). Using the chain rule (\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}). (\frac{dy}{du}=4u^{3}) and (\frac{du}{dx}=2x), so (\frac{dy}{dx}=4(x^{2}-1)^{3}\cdot2x=8x(x^{2}-1)^{3}).

Step2: Find where (\frac{dy}{dx}<0)

Set (\frac{dy}{dx}=8x(x - 1)^{3}(x + 1)^{3}<0). We can use a sign - chart. Test intervals:

• For (x\in(-\infty,- 1)), let (x=-2). Then (\frac{dy}{dx}=8\times(-2)\times((-2)^{2}-1)^{3}=8\times(-2)\times(3)^{3}<0).
• For (x\in(-1,0)), let (x =-\frac{1}{2}). Then (\frac{dy}{dx}=8\times(-\frac{1}{2})\times((-\frac{1}{2})^{2}-1)^{3}=8\times(-\frac{1}{2})\times(-\frac{3}{4})^{3}>0).
• For (x\in(0,1)), let (x=\frac{1}{2}). Then (\frac{dy}{dx}=8\times\frac{1}{2}\times((\frac{1}{2})^{2}-1)^{3}=8\times\frac{1}{2}\times(-\frac{3}{4})^{3}<0).
• For (x\in(1,\infty)), let (x = 2). Then (\frac{dy}{dx}=8\times2\times((2)^{2}-1)^{3}=8\times2\times(3)^{3}>0).

Answer:

((-\infty,-1)) and ((0,1))