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miss Akunina [59]
2 years ago
5

1.

Mathematics
1 answer:
Luba_88 [7]2 years ago
5 0

Hey there!

a = 5 ; b = 2

Option A.

a + b

= 5 + 2

= 7

Option B.

a - b

= 5 - 2

= 3

Option C.

2a + 3b

= 2(5) + 3(2)

= 10 + 6

= 16

Option D.

5a - b

= 5(5) - 2

= 25 - 2

= 23

Good luck on your assignment and enjoy your day!

~Amphitrite1040:)

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A farmer wants to fence in a rectangular plot of land adjacent to the north wall of his barn. No fencing is needed along the bar
Luba_88 [7]

The dimensions for the plot that would enclose the most area are a length and a width of 125 feet.

In this question we shall use the first and second derivative tests to determine the <em>optimal</em> dimensions of a rectangular plot of land. The perimeter (p), in feet, and the area of the rectangular plot (A), in square feet, of land are described below:

p = 2\cdot (w+l) (1)

A = w\cdot l (2)

Where:

  • w - Width, in feet.
  • l - Length, in feet.

In addition, the cost of fencing of the rectangular plot (C), in monetary units, is:

C = c\cdot p (3)

Where c is the fencing unit cost, in monetary units per foot.

Now we apply (2) and (3) in (1):

p = 2\cdot \left(\frac{A}{l}+l \right)

\frac{C}{c} = 2\cdot (\frac{A}{l}+l )

\frac{C\cdot l}{c} = 2\cdot (A+l^{2})

\frac{C\cdot l}{c}-2\cdot l^{2} = 2\cdot A

\frac{C\cdot l}{2\cdot c} - l^{2} = A (4)

We notice that fencing costs are directly proportional to the area to be fenced. Let suppose that cost is the <em>maximum allowable </em>and we proceed to perform the first and second derivative tests:

FDT

\frac{C}{2\cdot c}-2\cdot l = 0

l = \frac{C}{4\cdot c}

SDT

A'' = -2

Which means that length leads to a <em>maximum</em> area.

If we know that c = 8 and C = 4000, then the dimensions of the rectangular plot of land are, respectively:

l = \frac{4000}{4\cdot (8)}

l = 125\,ft

A = \frac{(4000)\cdot (125)}{2\cdot (8)} -125^{2}

A = 15625\,ft^{2}

w = \frac{15625\,ft^{2}}{125\,ft}

w = 125\,ft

The dimensions for the plot that would enclose the most area are a length and a width of 125 feet.

We kindly invite to check this question on areas: brainly.com/question/11952845

8 0
2 years ago
Which side of the equation 12.6v = 12 - 15.5v would you isolate the variable terms? Why?
lozanna [386]
The left side, because you only need to move one thing, and you don't need to subtract after that. (you can add, much more easier)

hope this helps
4 0
2 years ago
I'm havingn trouble in math. What's 3/4 - 2/5 in fraction form?
snow_tiger [21]
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6 0
2 years ago
Leroi and Sylvia both put $100 in a savings account. Leroi decides he will put in an additional $10 each week. Sylvia decides to
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Sylvia’s account

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7 0
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A ship is sailing due north. At a certain point the bearing of a lighthouse is N 42.8∘E and the distance is 10.5. After a while,
bija089 [108]

Answer:

How far did the ship travel between the two observations of the lighthouse  = 9.29

Step-by-step explanation:

the first step to answer this question is drawing the illustration as the attachment.

P is the ship, R is the light house and Q is the bearing.

PR is the distance between the ship and the light house, PR = 10.5

∠P = 42.8°, ∠Q = 59.7°

Thus, ∠R = 180° - ∠P  - ∠Q

               = 180° - 42.8°- 59.7°

               = 77.5°

PQ is the the distance of the ship moving. We can use the sinus equation

\frac{PR}{sin R} = \frac{PQ}{sin Q}

\frac{10.5}{sin 77.5°} = \frac{PQ}{sin 59.7°}

PQ = (\frac{10.5}{Sin 77.5°})(sin 59.7°)

     = 9.29

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