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ohaa [14]
4 years ago
15

Y=-x^2+20x-64 what is the max height

Mathematics
1 answer:
Inessa [10]4 years ago
8 0

Answer:

36

Step-by-step explanation:

The maximum height is the y-coordinate of the vertex

given a quadratic in standard form : ax² + bx + c : a ≠ 0

then the x-coordinate of the vertex is

x_{vertex} = - \frac{b}{2a}

y = - x² + 20x - 64 is in standard form

with a = - 1, b = 20 and c = - 64, hence

x_{vertex} = - \frac{20}{-2} = 10

substitute x = 10 into the equation for y

y = - (10)² + 20(10) - 64 = 36 ← max height


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The drama club sold 415 Manatee tickets to the school's play for $7.95 each 18 of the tickets were refunded estimate the tickets
nevsk [136]
Since you always round up when you estimate in the middle lets say theres 420 tickets. Each is 7.95, but we'll round it to 8.00. 420x8.=3,360. After that part we need to find about how much tickets were refunded. So we round 18 to 20 and each ticket is still 8$. 20x8=160. Finally in all, we do 3,360-160=3200. Its an overestimate because we rounded up. Im not sure what else youre asking but thats the basic answer.
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A ½in diameter rod of 5in length is being considered as part of a mechanical linkagein which it can experience a tensile loading
Evgesh-ka [11]

Answer:

a. Maximum Load = Force = 27085.09 N

b. Maximum Energy = 3.440 Joules

Step-by-step explanation:

Given

Rod diameter = ½in = 0.5in

Length = 5in

Young's modulus = 15.5Msi

By applying the 0.2% offset rule,

The maximum load the rod can hold before it gets to breaking point is given as follows by taking the strain as 0.2%

Young Modulus = Stress/Strain ------- Make Stress the Subject of Formula

Stress = Strain * Young Modulus

Stress = 0.2% * 15.5

Stress = 0.002 * 15.5

Stress = 0.031Msi

Calculating the area of the rod

Area = πr² or πd²/4

Area, A = 22/7 * 0.5^4 / 4

A = 22/7 * 0.25 / 4

A = 5.5/28

A = 0.1964in²

The maximum load that the rod would take before it starts to permanently elongate is given by

Force = Stress * Atea

Force = 0.031Msi * 0.1964in²

Force = 31Ksi * 0.1964in²

Force = 6.089Ksi in²

Force = 6.089 * 1000lbf

Force = 6089 lbf

1 lbf = 4.4482N

So, Force = 6089 * 4.4482N

Force = 27085.09 N

b.

Using Strain to Energy Formula

U = V×σ²/2·E

Where V = Volume

V = Length * Area

V = 5 in * 0.1964in²

V = 0.982in³

σ = Stress = 0.031Msi

= 0.031 * 1000Ksi

= 31Ksi

= 31 * 1000psi

= 31000psi

E = Young Modulus = 15.5Msi

= 15.5 * 1000Ksi

= 15.5 * 1000 * 1000psi

= 15500000psi

So,

Energy = 0.982 * 31000²/ ( 2 * 15500000)

Energy = 943,702,000/31000000

Energy = 30.442in³psi

------- Converted to ftlbf

Energy = 2.537 ftlbf

-------- Converted to Joules

Energy = 3.440 Joules

7 0
4 years ago
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