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Simora [160]
2 years ago
14

A valet makes $53 each day he works and approximately $7 in tips for each car he parks. if he wants to make at least $186 in one

day, at least how many cars does he need to park?
Mathematics
1 answer:
eimsori [14]2 years ago
5 0

Answer:

19 Cars

Step-by-step explanation:

Write a linear equation:

y=7x+53

Y = Total money

X = cars parked

If we want to earn 186 dollars a day then sub y for 186 and solve x:

186=7x+53

186-53=7x

133=7x

7x=133

x=133/7

x=19

This means that he will need to park 19 cars a day to earn a total of 186 dollars in one day.

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pogonyaev
F(x)=x^3-7x-6  Since I don't have the graph and this is not a perfect cube, I will have to rely on Newton :P

x-(f(x)/(dy/dx))

x-(x^3-7x-6)/(3x^2-7)

(2x^3+6)/(3x^2-7), letting x1=0

0, -6/7, -.988, -.9999, -.99999999999, -1

(x^3-7x-6)/(x+1)

x^2 r -x^2-7x-6
-x   r  -6x-6
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4 0
3 years ago
Read 2 more answers
Rafeeq bought a field in the form of a quadrilateral (ABCD)whose sides taken in order are respectively equal to 192m, 576m,228m,
Valentin [98]

Answer:

a. 85974 m²

b. 17,194,800 AED

c. 18,450 AED

Step-by-step explanation:

The sides of the quadrilateral are given as follows;

AB = 192 m

BC = 576 m

CD = 228 m

DA = 480 m

Length of a diagonal AC = 672 m

a. We note that the area of the quadrilateral consists of the area of the two triangles (ΔABC and ΔACD) formed on opposite sides of the diagonal

The semi-perimeter, s₁,  of ΔABC is found as follows;

s₁ = (AB + BC + AC)/2 = (192 + 576 + 672)/2 = 1440/2 = 720

The area, A₁, of ΔABC is given as follows;

Area\, of \, \Delta ABC = \sqrt{s_1\cdot (s_1 - AB)\cdot (s_1-BC)\cdot (s_1 - AC)}

Area\, of \, \Delta ABC = \sqrt{720 \times (720 - 192)\times  (720-576)\times  (720 - 672)}

Area\, of \, \Delta ABC = \sqrt{720 \times 528 \times  144 \times  48} = 6912·√(55) m²

Similarly, area, A₂, of ΔACD is given as follows;

Area\, of \, \Delta ACD= \sqrt{s_2\cdot (s_2 - AC)\cdot (s_2-CD)\cdot (s_2 - DA)}

The semi-perimeter, s₂,  of ΔABC is found as follows;

s₂ = (AC + CD + D)/2 = (672 + 228 + 480)/2 = 690 m

We therefore have;

Area\, of \, \Delta ACD = \sqrt{690 \times (690 - 672)\times  (690 -228)\times  (690 - 480)}

Area\, of \, \Delta ACD = \sqrt{690 \times 18\times  462\times  210} = \sqrt{1204988400} = 1260\cdot \sqrt{759} \ m^2

Therefore, the area of the quadrilateral ABCD = A₁ + A₂ = 6912×√(55) + 1260·√(759) = 85973.71 m² ≈ 85974 m² to the nearest meter square

b. Whereby the cost of 1 meter square land = 200 AED, we have;

Total cost of the land = 200 × 85974 = 17,194,800 AED

c. Whereby the cost of fencing 1 m = 12.50 AED, we have;

Total perimeter of the land = 576 + 192 + 480 + 228 = 1,476 m

The total cost of the fencing the land = 12.5 × 1476 = 18,450 AED

4 0
3 years ago
Any one know the answer it has to be turned in tomorrow morning
Veseljchak [2.6K]
17: 900 because 3,000 + 900 + 40 +7 is 3,947 or 3,947-3,000-40-7= 900
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schepotkina [342]
55% to 60%
The pattern of pellets within a 30-inch circle should be of a proper, even density to ensure a clean kill. The pattern should contain a sufficient percentage of the load, which should be at least 55% to 60%.
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Answer:

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