Plz plz , idk how to solve these. ; w; will give
i'd also appreciate a short step-by-step explanation if possible!

(perimeter of yard: 43.82m
area of yard: 119.1 m2)

let’s say we have access to planting 2 different crops, watermelons and grapes. we can find the revenue of a crop by the equation:

r = (p0 + px)(n0 - nx)

where r is the revenue, p0 is the starting price, n0 is the amount sold at the starting price, p is the price increase, n is the decrease in the amount sold per price increase, and x is the number of price increases.

crop a: watermelons have a starting price of $12.00 per m2 of area sold. at this price, you can sell 45 m2 worth. for every $2.00 increase, you will sell 1 m2 less. so the first task is to figure out how to put these values into the equation above and turn it into a quadratic function. then you will need to find the value of x to make the revenue as large as possible. once you know x, the number of price increases, you can then find what you should price your watermelons per m2.

equation: r =

number of price increases: x =

price of watermelon per m^2 =

okay, now we can do the same thing with the grapes, just to give you that extra practice with quadratic functions. the starting price of grapes per m² is $8.00, at which you can sell 60 m² worth. for every $3.00 increase in price, you sell 3 m² less. so use the same method you just used to find the price of grapes per m² of area to maximize revenue.

equation: r =

number of price increases: x =

price of grape per m^2 =

okay, so now, all in all, we have built our farm and found what to price each of our crops at. now let’s find what amount of each crop to grow and find our profits. getting right into it, we have the following cost equations:

cw = $20.00 * w + $50.00

cg = $10.00 * g + $40.00

where cw is the total cost of planting and growing watermelons, cg is the total cost of planting and growing grapes, w is the area of watermelons planted, and g is the area of grapes planted. now that we know our cost, we can finally find our profits. profit is equal to the price per unit area, multiplied by the area sold, subtracted by the total cost planting and growing them, or, profit = price * area - cost. but at this point, we still don’t know the area of each crop in our field. but we do know a couple of things to figure it out. we know the total area of our farm, found in the first part. which we can say is, area = g + w and we were also only given $1590 to plant the crops. so, $1590.00= cw+cg i’ll you simplify this a bit:

$1590.00=$20.00*w+$50.00+$10.00*g+$ 40.00 $1590.00=$20.00*w+$10.00*g+$90.00
so now we have enough information to solve for the areas of watermelons and grapes in your field. here’s some room to do that. hint: use either substitution or graph the two equations to find an intersection.

area of watermelon to plant:

area of grapes to pant:

profit of the crops:

56.6

step-by-step explanation:

the ratio of the perimeters is equal to the scale.

the ratio of the areas is equal to the square of the scale.

34 / 20 = k

a / 19.6 = k²

substituting:

a / 19.6 = (34 / 20)²

a = 19.6 (34 / 20)²

a ≈ 56.6

lockers 1, 4, 9, 16, 25, 40, and 49 are open

i ran through it by the other answer is wrong

here we can see that the storage increases

at certain times of the year - during the

warmer months, when the gas is not much

used for heating and reserves can be

increased - and then decreases during the

colder months because the supply is being

used. it is possible to model this with a

relatively simple trigonometric graph, but

first we examine the simplest trig curves.

065/065.html

the basics

the simplest sine and cosine curves can be generated by plotting the trig-function values

against the angles used in each function. the values can be found in the chart below:

ø (˚) 0 30 45 60 90 120 135 150 180 210 225 240 270 300 315 330 360

sin ø

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0 1

2

2

2 3

2 1 3

2 2

2 1

2

0 − 1

2

− 2

2 − 3

2 − 1 − 3

2 − 2

2 − 1

2

0

cos ø

€

1 3

2 2

2 1

2 0 − 1

2

− 2

2 − 3

2 − 1 − 3

2 − 2

2 − 1

2

0 1

2 2

2 3

2 1

first we plot the sine data, letting the x-axis represent the angles and the y-axis the sine

values. the result is the scatter plot below.

based on just these 17 data points, we can see

that the function is following a straightforward

pattern - the sine values are increasing to 1,

decreasing through the same values to -1, then

returning to zero. when we connect the dots, we

get the periodic curve y = sin (x) below.

this “sine curve,” with equation

y = sin(x), is the simplest of trig

graphs. the curve rises and falls by

one unit from its “axis of

oscillation” - in this case, the x-axis

- and after we cycle through 360˚ it

appears we are ready to go through

the same y-values again. we can see

this same pattern in the cosine

function on the following page.

each sine and cosine curve has two basic characteristics:

amplitude - distance the curve rises above, and falls below, its axis of oscillation (marked by

the vertical arrows on the cosine graph). the formula is

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amp = max− min

2 .

period - the length of one cycle (marked by the horizontal arrows on the cosine graph).

this can be calculated by measuring the distance between any successive maximum

points (or minimum points), or by doubling the horizontal distance between a

minimum and the next maximum.

for our first two curves, then, the amplitude is 1 and the period is 360˚ (or 2! if we are in radians).

adjusting the amplitude and frequency: y = a sin(bx), y = a cos(bx)

now that we have our “base” amplitude and frequency . . let’s change them. we can adjust both

characteristics by introducing the parameters a and b into the equations. the graph below shows the

curves y = a sin(x) for a = 1,2,3, and -0.5, with the original curve y = sin(x) being the thickest line.

clearly the amplitude of the curves has changed;

when a = 2, the amplitude is 2; when a = 3, the

amplitude is 3; and when a = -0.5 the amplitude is

0.5 (since amplitude, a distance, must be positive).

however, this last curve has also been flipped upside

down (or “reflected in the x-axis”). since the

amplitude must be positive even when a itself is

negative, we have the formula

amplitude = | a |.

the relationship between the b-value and the

period is not so straightforward. on the right

we have y = cos(bx) for b = 1,3, and 0.5. (the

thickest line is again our “base” curve.) when

b=3, the period decreases to 120˚; when

b=0.5, the period increases to 720˚ (so only

half the curve is shown here). we can describe

this reciprocal relationship with the formula

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b = 360

period or

€

b = 2π

period ,

using the latter if we are working in radians.

shifting horizontally and vertically: y = sin(x-h) + k, y = cos(x-h) + k

to shift (or “translate”) a curve along either the x- or y-axis (or both), we introduce the parameters

h and k. below we have the graph of y = sin (x - 30) + 2 and y = sin(x). (note that we have returned

to an amplitude of 1 and period of 360˚). our original

curve y = sin(x) had its peak at the point (90,1); the

shifted curve has its corresponding peak at (120, 3). in

the same way, the original curve had a minimum value

at (270, -1); now the minimum is at (300,1). and

finally, the axis of oscillation has shifted vertically

from the line x = 0 to the line x =2. in other words, our

new curve has been created by shifting y = sin(x) right

30 and u p 2. and, in general, we can create the

following shifts:

horizontal shift = h (left if h < 0, right if h > 0)

vertical shift/new axis of oscillation = k.

modelling data: y = a sin[b(x-h)] + k, y = a cos[b(x-h)] + k

we have seen how the basic sine and cosine curves can be adjusted: by changing the amplitude

and period, and by shifting the curves horizontally and vertically. a summary is below.

amplitude = | a | vertical shift/axis of oscillation = k

period =

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360

b

or

€

2π

b