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Mazyrski [523]
3 years ago
5

A cable runs from the top of a building to a point on the ground 60.3 feet from the base of the building. If the cable makes and

angle of 33.7˚ with the ground (see figure), find the following: a. Find the height of the building. Round your answer to the nearest foot. b. Without using your answer from part a, find the length of the cable. Round your answer to the nearest foot.
Mathematics
1 answer:
cupoosta [38]3 years ago
3 0
Part A:

Given that the <span>cable runs from the top of a building to a point on the ground 60.3 feet from the base of the building and that the </span>cable makes and angle of 33.7˚ with the ground.

Let the height of the building be h, then

\tan33.7^o= \frac{h}{60.3}  \\  \\ \Rightarrow h=60.3\tan33.7^o \\  \\ \approx40\, feet.



Part B:

Given that the cable runs from the top of a building to a point on the ground 60.3 feet from the base of the building and that the cable makes and angle of 33.7˚ with the ground.

Let the length of the building be l, then

\cos33.7^o= \frac{60.3}{l}  \\  \\ \Rightarrow l=\frac{60.3}{\cos33.7^o} \\  \\ \approx72\, feet.
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dmitriy555 [2]

Answer:

The distance between (-9, 8) and (-3, -1) will be:

\:d=3\sqrt{13} units

Step-by-step explanation:

Given the points

  • (-9, 8)
  • (-3, -1)

Finding the distance between (-9, 8) and (-3, -1)

d\:=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}

   =\sqrt{\left(-3-\left(-9\right)\right)^2+\left(-1-8\right)^2}

   =\sqrt{\left(-3+9\right)^2+\left(-1-8\right)^2}

   =\sqrt{36+81}

   =\sqrt{117}

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Therefore, the distance between (-9, 8) and (-3, -1) will be:

\:d=3\sqrt{13} units

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3 years ago
Can someone help me on this
barxatty [35]

A) as the exponent decreases by one, the number is divided by 5

B) as the exponent decreases by one, the number is divided by 4

C) as the exponent decreases by one, the number is divided by 3

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    4⁻¹ = 1 ÷ 4 = \frac{1}{4}

    4⁻² = \frac{1}{4} ÷ 4 = \frac{1}{16}

E) 3⁰ = 3 ÷ 3 = 1

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Answer:

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At time, 20 seconds, the mass remaining is 12 grams

At time, 30 seconds, the mass remaining is 6 grams

At time, 40 seconds, the mass remaining is 3 grams

At time, 50 seconds, the mass remaining is 1.5 grams

Step-by-step explanation:

Given;

initial mass of the radioactive element  = 48 grams

half life, t = 10 seconds

time of decay (s)                          remaining mass of the  radioactive element

0 ------------------------------------------------- 48 grams

10------------------------------------------------- 24 grams

20 ------------------------------------------------- 12 grams

30 -------------------------------------------------- 6 grams

40 --------------------------------------------------- 3 grams

50 ----------------------------------------------------- 1.5 grams

60------------------------------------------------------0.75 grams

Thus;

At time, 10 seconds, the mass remaining is 24 grams

At time, 20 seconds, the mass remaining is 12 grams

At time, 30 seconds, the mass remaining is 6 grams

At time, 40 seconds, the mass remaining is 3 grams

At time, 50 seconds, the mass remaining is 1.5 grams

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Answer:

Part C

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What can you say about AB, about the inscribed angle of a circle, and include your reasoning in your response.

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