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2 years ago

The passage states that the farm managers react to

1 answer:

8 0

The passage indicates that the farm's managers reacted to Nawab's receipt of a motorcycle with disgust. The correct answer is A.

Given a passage shown below in the attached pictures.

Option A is the best answer. The passage says that Nawab's new motorcycle led to "the disgust of the ranchers" (line 7 ).

Options other than A are B, C, and D are incorrect because the passage specifically states that Nawab's new motorcycle resulted in "the disgust of the ranchers", not happiness. their happiness (option B), their jealousy (option C) or their indifference (option D). The passage is therefore given that the managers of the farm reacted to Nawab's receipt of a motorcycle with disgust.

Learn more about the passage from here brainly.com/question/12555695

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Help please help me out please

maxonik [38]

Answer:

$-\frac{28}{3}$

Step-by-step explanation:

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2 years ago

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Multiply (x - 4)(2x + 3) using the foil method show the foil method products

AVprozaik [17]

(x - 4)(2x + 3)

Using foil:

2x^2 + 3x - 8x - 12

Simplify.

2x^2 - 5x - 12 is the product after using FOIL.

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3 years ago

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A rancher wishes to build a fence to enclose a 2250 square yard rectangular field. Along one side the fence is to be made of hea

Bess [88]

Answer:

The least cost of fencing for the rancher is $1200

Step-by-step explanation:

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

Let C the total cost of the rectangular field.

The side made of heavy duty material of length of x costs 16 dollars a yard. The three sides not made of heavy duty material cost $4 per yard, their side lengths are x, y, y. Thus

$C=4x+4y+4y+16x\\C=20x+8y$

We know that the total area of rectangular field should be 2250 square yards,

$x\cdot y=2250$

We can say that $y=\frac{2250}{x}$

Substituting into the total cost of the rectangular field, we get

$C=20x+8(\frac{2250}{x})\\\\C=20x+\frac{18000}{x}$

We have to figure out where the function is increasing and decreasing. Differentiating,

$\frac{d}{dx}C=\frac{d}{dx}\left(20x+\frac{18000}{x}\right)\\\\C'=20-\frac{18000}{x^2}$

Next, we find the critical points of the derivative

$20-\frac{18000}{x^2}=0\\\\20x^2-\frac{18000}{x^2}x^2=0\cdot \:x^2\\\\20x^2-18000=0\\\\20x^2-18000+18000=0+18000\\\\20x^2=18000\\\\\frac{20x^2}{20}=\frac{18000}{20}\\\\x^2=900\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{900},\:x=-\sqrt{900}\\\\x=30,\:x=-30$

Because the length is always positive the only point we take is $x=30$ . We thus test the intervals $(0, 30)$ and $(30, \infty)$

$C'(20)=20-\frac{18000}{20^2} = -25 < 0\\\\C'(40)= 20-\frac{18000}{20^2} = 8.75 >0$

we see that total cost function is decreasing on $(0, 30)$ and increasing on $(30, \infty)$ . Therefore, the minimum is attained at $x=30$ , so the minimal cost is

$C(30)=20(30)+\frac{18000}{30}\\C(30)=1200$