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In-s [12.5K]
2 years ago
6

4/3y=48solve for ysimplify answer as much as posible

Mathematics
1 answer:
Helga [31]2 years ago
6 0
<h2>If you mean :</h2>

\frac{4}{3y}  = 48 \\

Inverse both sides

\frac{3y}{4}  =  \frac{1}{48}  \\

Multiply both sides by 4

4 \times  \frac{3y}{4}  = 4 \times  \frac{1}{48}  \\

3y =  \frac{4}{48}  \\

3y =  \frac{4 \times 1}{4 \times 12}  \\

3y =  \frac{1}{12}  \\

Divide both sides by 3

3y \div 3 =  \frac{1}{12}  \div 3 \\

y =  \frac{1}{12}  \times  \frac{1}{3}  \\

y =  \frac{1}{36}  \\

_____________________________________________

<h2>If u mean : </h2>

\frac{4}{3} y = 48 \\

Multiply both sides by the 3

3 \times \frac{4}{3} y = 3 \times 48 \\

4y = 3 \times 48

Divide both sides by 4

4y  \div 4 = 3 \times 48 \div 4

y =  \frac{48 \times 3}{4}  \\

y =  \frac{4 \times 12 \times 3}{4 \times 1}  \\

y = 12 \times 3

y = 36

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  • Part a)  \bf{\sqrt{x^2+(x^2-3)^2}
  • Part b)  3
  • Part c)   2.24
  • Part d)  1.58

============================================================

Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

We apply the distance formula to say...

d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying d = \sqrt{x^2 + (x^2-3)^2} is the same as writing d = sqrt(x^2-(x^2-3)^2)

-------------------------------------------

Part (b)

Plug in x = 0 and you should find the following

d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

-------------------------------------------

Part (c)

Repeat for x = 1

d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\

-------------------------------------------

Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.

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