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Nadya [2.5K]
3 years ago
14

Each book in a store costs $8, and each pen costs $4. If you want to spend exactly $32, write the equation, in standard form, th

at represents modeling this situation. Let B represent the number of books you buy, and P represent the number of pens you buy.
Mathematics
1 answer:
Lena [83]3 years ago
5 0

Answer:

Step-by-step explanation:

Let B represent the number of books that you buy, and let P represent the number of pens that you buy.

If each book in a store costs $8 and each pen costs $4, it means that the cost of B books and P pens would be expressed as

8B + 4P

If you want to spend exactly $32, then the equation, in standard form, that represents modeling this situation is

8B + 4P = 32

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Anna35 [415]

Answers:

  • 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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2 years ago
I need help with this, pls help
Alex

Answer:

Step-by-step explanation:

why

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Answer:

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Step-by-step explanation:

The distance from a point (m, n ) to the line Ax + By + C = 0 is given by

d = \frac{|Am+Bn+C|}{\sqrt{A^2+B^2} }

For the point (- 1, 3 ) , with m = - 1 and n = 3

3x - 4y = 10 ( subtract 10 from both sides )

3x - 4y - 10 = 0

with A = 3 , B = - 4 , C = - 10

d = \frac{|3(-1)+(-4)3-10|}{\sqrt{3^2+(-4)^2} }

   = \frac{|-3-12-10|}{9+16}

   = \frac{|-25|}{\sqrt{25} }

    = \frac{25}{5}

     = 5

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