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professor190 [17]
3 years ago
5

The function L = 0.8T^2 models the relationship between L, the length in feet of a pendulum, and T, the period in seconds of the

pendulum. Which value is closest to the period in seconds for a pendulum that is 30 ft long?
a 5.4s
b 4.9s
c 6.8s
d 6.1s
Mathematics
1 answer:
Semmy [17]3 years ago
7 0
The correct option is "d".
Given that L = 0.8T²
length of pendulum = 30ft

L= 0.8T²
30 = 0.8T²
T² = 30 / 0.8
T² = 37.5
T = √37.5 = 6.1 seconds
<span>So, 6.1 is the closest to the period in seconds for a pendulum that is 30 ft long.</span>
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Given:

Length of the ladder = 9.85 m

The height to the top of the ladder is 5 m more than the distance between the wall and the foot of the ladder.

To find:

The height to the top and the distance between the wall and the foot of the ladder.​

Solution:

let x be the distance between the wall and the foot of the ladder. Then the height to the top of the ladder is (x+5).

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Hypotenuse^2=Base^2+Perpendicular^2

In the given situation, hypotenuse is the length of ladder, i.e., 9.85 m. The base is x m and the height is (x+5) m.

Using the Pythagoras theorem, we get

(9.85)^2=x^2+(x+5)^2

97.0225=x^2+x^2+10x+25

0=2x^2+10x+25-97.0225

0=2x^2+10x-72.0225

Here, a=2, b=10,c=-72.0225. Using the quadratic formula, we get

x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}

x=\dfrac{-10\pm \sqrt{(10)^2-4(2)(-72.0225)}}{2(2)}

x=\dfrac{-10\pm \sqrt{676.18}}{4}

Approximating the value, we get

x=\dfrac{-10\pm 26}{4}

x=\dfrac{-10+26}{4},\dfrac{-10-26}{4}

x=\dfrac{16}{4},\dfrac{-36}{4}

x=4,-9

Distance cannot be negative so x\neq -9.

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A movie theater has a seating capacity of 329. The theater charges $5.00 for children, $7.00 for students, and $12.00 of adults.
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Answer:

170 children

74 students

85 adults

Step-by-step explanation:

Given

Let:

C = Children; S = Students; A = Adults

For the capacity, we have:

C + S + A = 329

For the tickets sold, we have:

5C + 7S + 12A = 2388

Half as many as adults are children implies that:

A = \frac{C}{2}

Required

Solve for A, C and S

The equations to solve are:

C + S + A = 329 -- (1)

5C + 7S + 12A = 2388 -- (2)

A = \frac{C}{2} -- (3)

Make C the subject in (3)

C = 2A

Substitute C = 2A in (1) and (2)

C + S + A = 329 -- (1)

2A + S + A = 329

3A + S = 329

Make S the subject

S = 329 - 3A

5C + 7S + 12A = 2388 -- (2)

5*2A + 7S + 12A = 2388

10A + 7S + 12A = 2388

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Substitute S = 329 - 3A

7(329 - 3A) + 22A = 2388

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2303 +A = 2388

Solve for A

A = 2388 - 2303

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C = 2 * 85

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Recall that: S = 329 - 3A

S = 329 - 3 * 85

S = 329 - 255

S = 74

Hence, the result is:

C = 170

S = 74

A = 85

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