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Olegator [25]
3 years ago
11

- Alto and Sheryl both worked hard over the summer. Together they earned a total of $475. Sheryl earned $25 more than Alto. (a)

Write a system of equations for the situation. Use s for the amount Sheryl earned and a for the amount Alto earned. a (b) Graph the equations of the system on the graph provided on the next page. a. Note: Insert->Shape will allow you to add lines, draw dots, and otherwise insert shapes in both Microsoft Word and LibreOffice. The graph should be big enough for you to easily plot points using this method. b. If you do not want to graph using the drawing tools built into Microsoft Word or LibreOffice you may also print out, graph by hand, and scan the page. Please note it's perfectly okay to upload multiple files. (c) Use Elimination or Substitution to solve the system. (d) Use your graph to estimate how much each person earned, check your work with part C, and explain your results in complete sentences. Answer:
pls hurry its due tomorrow​
Mathematics
2 answers:
Ksenya-84 [330]3 years ago
4 0

Answer:

s = 250

a = 225

Step-by-step explanation:

Because they both earned a total of 475 dollars, the total amount of the two is 475. Because a represents Alto's earnings, and s represents Sheryl's earnings, we have a+s=475

Sheryl also earned 25 more dollars than Alto, so the value of s is 25 larger than a: s=a+25

By graphing, we can let s = y and a = x, so we have the following equations:

x + y=475\\y=x+25

Plugging this into Desmos, we have this: (look at screenshot uploaded)

We see that y, or Sheryl's amount is around 250, and x, or Alto's amount is in the middle of 200 and 250, roughly about 225.

Now, we can solve for a and s. Because the value of s is equal to a + 25, we can represent s as a + 25 in the first equation: a + (a + 25) = 475. Combining like terms, we have 2a+25=475, and 2a = 450. Dividing, we have a = 225, and since s is 25 more than 225, we have s = 250

The amount we estimated was around what we solved for. We found that the solution, or the value where both cases are true is around 250 and 225 for Sheryl and Alto's earnings.

trasher [3.6K]3 years ago
3 0
475 = s + a
s = a + 25
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A(t)=.892t^3-13.5t^2+22.3t+579 how to solve this
Minchanka [31]

Answer:

t = (5 ((446 sqrt(3188516012553) - 827891226)^(1/3) - 204292 (-1)^(2/3) (3/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3)) + 1125/223 or t = 1125/223 - (5 ((-2)^(1/3) (223 sqrt(3188516012553) - 413945613)^(1/3) - 204292 (-3/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3)) or t = 1125/223 - (5 ((827891226 - 446 sqrt(3188516012553))^(1/3) + 204292 (3/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3))

Step-by-step explanation:

Solve for t over the real numbers:

0.892 t^3 - 13.5 t^2 + 22.3 t + 579 = 0

0.892 t^3 - 13.5 t^2 + 22.3 t + 579 = (223 t^3)/250 - (27 t^2)/2 + (223 t)/10 + 579:

(223 t^3)/250 - (27 t^2)/2 + (223 t)/10 + 579 = 0

Bring (223 t^3)/250 - (27 t^2)/2 + (223 t)/10 + 579 together using the common denominator 250:

1/250 (223 t^3 - 3375 t^2 + 5575 t + 144750) = 0

Multiply both sides by 250:

223 t^3 - 3375 t^2 + 5575 t + 144750 = 0

Eliminate the quadratic term by substituting x = t - 1125/223:

144750 + 5575 (x + 1125/223) - 3375 (x + 1125/223)^2 + 223 (x + 1125/223)^3 = 0

Expand out terms of the left hand side:

223 x^3 - (2553650 x)/223 + 5749244625/49729 = 0

Divide both sides by 223:

x^3 - (2553650 x)/49729 + 5749244625/11089567 = 0

Change coordinates by substituting x = y + λ/y, where λ is a constant value that will be determined later:

5749244625/11089567 - (2553650 (y + λ/y))/49729 + (y + λ/y)^3 = 0

Multiply both sides by y^3 and collect in terms of y:

y^6 + y^4 (3 λ - 2553650/49729) + (5749244625 y^3)/11089567 + y^2 (3 λ^2 - (2553650 λ)/49729) + λ^3 = 0

Substitute λ = 2553650/149187 and then z = y^3, yielding a quadratic equation in the variable z:

z^2 + (5749244625 z)/11089567 + 16652679340752125000/3320419398682203 = 0

Find the positive solution to the quadratic equation:

z = (125 (223 sqrt(3188516012553) - 413945613))/199612206

Substitute back for z = y^3:

y^3 = (125 (223 sqrt(3188516012553) - 413945613))/199612206

Taking cube roots gives (5 (223 sqrt(3188516012553) - 413945613)^(1/3))/(223 2^(1/3) 3^(2/3)) times the third roots of unity:

y = (5 (223 sqrt(3188516012553) - 413945613)^(1/3))/(223 2^(1/3) 3^(2/3)) or y = -(5 (-1/2)^(1/3) (223 sqrt(3188516012553) - 413945613)^(1/3))/(223 3^(2/3)) or y = (5 (-1)^(2/3) (223 sqrt(3188516012553) - 413945613)^(1/3))/(223 2^(1/3) 3^(2/3))

Substitute each value of y into x = y + 2553650/(149187 y):

x = (5 ((223 sqrt(3188516012553) - 413945613)/2)^(1/3))/(223 3^(2/3)) - 510730/223 (-1)^(2/3) (2/(3 (413945613 - 223 sqrt(3188516012553))))^(1/3) or x = 510730/223 ((-2)/(3 (413945613 - 223 sqrt(3188516012553))))^(1/3) - (5 ((-1)/2)^(1/3) (223 sqrt(3188516012553) - 413945613)^(1/3))/(223 3^(2/3)) or x = (5 (-1)^(2/3) ((223 sqrt(3188516012553) - 413945613)/2)^(1/3))/(223 3^(2/3)) - 510730/223 (2/(3 (413945613 - 223 sqrt(3188516012553))))^(1/3)

Bring each solution to a common denominator and simplify:

x = (5 ((446 sqrt(3188516012553) - 827891226)^(1/3) - 204292 (-1)^(2/3) (3/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3)) or x = -(5 ((-2)^(1/3) (223 sqrt(3188516012553) - 413945613)^(1/3) - 204292 ((-3)/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3)) or x = -(5 ((827891226 - 446 sqrt(3188516012553))^(1/3) + 204292 (3/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3))

Substitute back for t = x + 1125/223:

Answer: t = (5 ((446 sqrt(3188516012553) - 827891226)^(1/3) - 204292 (-1)^(2/3) (3/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3)) + 1125/223 or t = 1125/223 - (5 ((-2)^(1/3) (223 sqrt(3188516012553) - 413945613)^(1/3) - 204292 (-3/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3)) or t = 1125/223 - (5 ((827891226 - 446 sqrt(3188516012553))^(1/3) + 204292 (3/(413945613 - 223 sqrt(3188516012553)))^(1/3)))/(223 6^(2/3))

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Helen [10]
6 : 10 = 3 : 5
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A. 18 : 20 = 9 : 10 (incorrect)
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D. 50 : 30 = 5 : 3 (incorrect)

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Answer:

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kati45 [8]

Answer:

Okay. At noon, the hour and minute hand overlap. First, let’s calculate the speed of the hands.

The minute hand rotates once around the clock every hour. It goes 360 degrees in 60 minutes, so it rotates 360/60 or 6 degrees per minute. The hour hand goes once around the clock every 12 hours. Because there are 60 minutes in an hour, it goes 360 degrees 12*60 or 720 minutes. This means it travels at a speed of 360/720 or 0.5 degrees per minute.

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