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Alik [6]
3 years ago
5

Can someone help me? I attached the question on here, I could really use some help.

Mathematics
1 answer:
frosja888 [35]3 years ago
3 0
I'd say the bottom answer is the best answer
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Help help help help please this is due soon
IgorLugansk [536]

Answer: (2,-1)

Step-by-step explanation:

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3 years ago
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A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single-strand electric f
likoan [24]

Step-by-step explanation:

Let x be the length and y be the width of the rectangular plot.

The plot is bounded on one side by a river and on the other three sides by a single-strand electric fence. It means,

x+2y = 1500

x = 1500 - 2y ....(1)

We know that the area of a rectangular plot is given by :

A = xy ....(2)

Put the value of x from equation (1) in (2)

A=(1500-2y)y\\\\A=1500y-2y^2 .....(3)

For largest area, differentiate above area equation wrt y.

\dfrac{dA}{dy}=\dfrac{d}{dy}(1500y-2y^2)\\\\=1500-4y\\\\\text{Put}\ \dfrac{dA}{dy}=0\\\\1500-4y=0\\\\y=\dfrac{1500}{4}\\\\=375

Put the value of y in equation (1).

x = 1500-2(375)

= 750 m

Put the value of y in equation (3).

A =1500(375)-2(375)^2\\A=281250\ m^2

Hence, the largest area is 281250 m² and its dimensions are 750 m and 375 m.

3 0
3 years ago
HELP!! Algebra help!! Will give stars thank u so much <333
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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3 years ago
The given line passes through the points (-1,-1) and (4.1).
ira [324]

Answer:

(-1,-1) and (4.1).

                                     1+1

slope of the line: m= ------- = 2/5

                                     4+1

the slope of the perpendicular line:  -5/2  and   point  (-1,3)

(y-3) = -5/2(x+1)

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3 years ago
Find the distance between the points (18,20) and (8,13)
ExtremeBDS [4]
2 and 5...................... hope I helped
7 0
3 years ago
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