Aconstruction crew is lengthening a road. let l be the total length of the road (in miles). let d be the number of days the crew has worked. suppose that l=2d + 400 gives l as a function of d. the crew can work for at most 80 days. identify the correct description of the values in both the domain and tange of the function. then, for each, choose the most appropriate set of values. set of values o number of days the crew has worked o length of the road (in miles) o number of days the crew has worked o length of the road (in miles) domain range the set of all real numbers from 0 to 80 the set of all real numbers from 400 to 560 the set of all real numbers from 2 to 400 (so. 81, 82. 83. 560 (o. 1. 2. 400) 2. 4. 6, 8.

Set of Values for Domain = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80}

Set of Values for Range = {402, 404, 406, 408, 410, 412, 414, 416, 418, 420, 422, 424, 426, 428, 430, 432, 434, 436, 438, 440, 442, 444, 446, 448, 450, 452, 454, 456, 458, 460, 462, 464, 466, 468, 470, 472, 474, 476, 478, 480, 482, 484, 486, 488, 490, 492, 494, 496, 498, 500, 502, 504, 506, 508, 510, 512, 514, 516, 518, 520, 522, 524, 526, 528, 530, 532, 534, 536, 538, 540 542, 544, 546, 548 550, 552, 554, 556, 558, 560}.

Step-by-step explanation:

Solution...

Assume that there is a link for the road between your house and the market the road maintenance team is meant to repair the road.

now the system of work is that the length of the road they will work upon is equal to 2 times the amount of days + 400.

This is to say that originally there is 400 miles of the road, and once they work in a day they add 2 more miles to the original 400. if they work in 2 days they add 4 more miles to the original 400. If they work in 3 days they add 6 more miles to the original 400.

So L = 2(D) + 400.

Answer 1.

description of the values in both the domain and range of a function.

the domain is the set of all values for which the function is defined.

this means that the domain is all numbers between 0 to 80.

Why do we begin from 0, because when you put 0 into the equation you have, L = 2(0) + 400, and this is equal to 2 x 0 + 400 or 400 + 0 (since all numbers multiplied by 0 gives 0).

Why is 80 the maximum, this is because they work for at most 80 days.

So the domain is

D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80}

And the range is the set of all values that L (which is the function) takes, to get this you have to calculate the days from day 0 to day 80.

The range for the function is...

L = {402, 404, 406, 408, 410, 412, 414, 416, 418, 420, 422, 424, 426, 428, 430, 432, 434, 436, 438, 440, 442, 444, 446, 448, 450, 452, 454, 456, 458, 460, 462, 464, 466, 468, 470, 472, 474, 476, 478, 480, 482, 484, 486, 488, 490, 492, 494, 496, 498, 500, 502, 504, 506, 508, 510, 512, 514, 516, 518, 520, 522, 524, 526, 528, 530, 532, 534, 536, 538, 540 542, 544, 546, 548 550, 552, 554, 556, 558, 560}.

So how did we get all these. Using the figures registered as D to get the L.

Before solving, remember that, (HD is H•D or H × D,

2D is 2•D or 2 × D,

29F is 29•F or 29 × F... When numbers are used to multiply letters which stand as a varying value like X, Y, D etc, then they are written as 2x or 2•x or 2(x). They all mean multiply 2 by x)

So, i.e.

L = 2(D) + 400.

When it's day 1, D is 1

L = 2(1) + 400 which equals 2 + 400 = 402. So for 1 day, the crew has worked 402 miles.

On Day 2, D is 2

L = 2(2) + 400 which equals 4 + 400 = 404. So for 2 days, the crew has worked 404 miles.

On Day 3, D is 3

L = 2(3) + 400 which equals 6 + 400 = 406. So for 3 days, the crew has worked 406 miles.

On Day 4, D is 4

L = 2(4) + 400 which equals 8 + 400 = 408. So for 4 days, the crew has worked 408 miles.

On Day 5, D is 5

L = 2(5) + 400 which equals 10 + 400 = 410. So for 5 days, the crew has worked 410 miles.

On Day 6, D is 6

L = 2(6) + 400 which equals 12 + 400 = 412. So for 6 days, the crew has worked 412 miles.

...

...

...

For 7 days, the crew has worked 414 miles.

For 8 days, the crew has worked 416 miles.

For 9 days, the crew has worked 418 miles.

For 10 days, the crew has worked 420 miles.

For 20 days, the crew has worked 440 miles.

For 30 days, the crew has worked 460 miles.

For 40 days, the crew has worked 480 miles.

For 50 days, the crew has worked 500 miles.

For 60 days, the crew has worked 520 miles.

For 70 days, the crew has worked 540 miles.

For 75 days, the crew has worked 550 miles.

...

...

...

On Day 79, D is 79

L = 2(79) + 400 which equals 158 + 400 = 558. So for 79 days, the crew has worked 558 miles..

On Day 80, D is 80

L = 2(80) + 400 which equals 160 + 400 = 560. So for 80 days, the crew has worked 560 miles.

That is how the domain and range is determined.

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt