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larisa86 [58]
3 years ago
7

A firework is launched at the rate of 10 feet per second from a point on the ground 50 feet from an observer. to 2 decimal place

s in radians per second, find the rate of change of the angle of elevation when the firework is 40 feet above the ground. type your answer in the space below. if your answer is a number less than 1, place a leading "0" before the decimal point (ex: 0.35).

Mathematics
1 answer:
Kazeer [188]3 years ago
5 0

The rate of change of the angle of elevation when the firework is 40 feet above the ground is 0.12 radians/second.

First we will draw a right angle triangle ΔABC, where ∠B = 90°

Lets, assume the height(AB) = h and base(BC)= x

If the angle of elevation, ∠ACB = α, then

tan(α) = \frac{AB}{BC} = \frac{h}{x}

Taking inverse trigonometric function, α = tan⁻¹ (\frac{h}{x}) .............(1)

As we need to find the rate of change of the angle of elevation, so we will differentiate both sides of equation (1) with respect to time (t) :

\frac{d\alpha}{dt}=[\frac{1}{1+ \frac{h^2}{x^2}}]*(\frac{1}{x})\frac{dh}{dt}

Here, the firework is launched from point B at the rate of 10 feet/second and when it is 40 feet above the ground it reaches point A,

that means h = 40 feet and \frac{dh}{dt} = 10 feet/second.

C is the observer's position which is 50 feet away from the point B, so x = 50 feet.

\frac{d\alpha}{dt}= [\frac{1}{1+ \frac{40^2}{50^2}}] *\frac{1}{50} *10\\ \\ \frac{d\alpha}{dt} = [\frac{1}{1+\frac{16}{25}}] *\frac{1}{5}\\ \\ \frac{d\alpha}{dt} = [\frac{25}{41}] *\frac{1}{5}\\   \\ \frac{d\alpha}{dt}= \frac{5}{41} =0.1219512

= 0.12 (Rounding up to two decimal places)

So, the rate of change of the angle of elevation is 0.12 radians/second.

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butalik [34]

Answer:

So the z-scores that separate the unusual IQ scores from those that are​ usual are Z = -2 and Z = 2.

The IQ scores that separate the unusual IQ scores from those that are​ usual are 84 and 148.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

\mu = 116, \sigma = 16

What are the z scores that separate the unusual IQ scores from those that are​ usual?

If Z<-2 or Z > 2, the IQ score is unusual.

So the z-scores that separate the unusual IQ scores from those that are​ usual are Z = -2 and Z = 2.

What are the IQ scores that separate the unusual IQ scores from those that are​ usual?

Those IQ scores are X when Z = -2 and X when Z = 2. So

Z = -2

Z = \frac{X - \mu}{\sigma}

-2 = \frac{X - 116}{16}

X - 116 = -2*16

X = 84

Z = 2

Z = \frac{X - \mu}{\sigma}

2 = \frac{X - 116}{16}

X - 116 = 2*16

X = 148

The IQ scores that separate the unusual IQ scores from those that are​ usual are 84 and 148.

8 0
3 years ago
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What point would A' be after a dilation with a scale factor of 2 centered at (-7,-2)?​
musickatia [10]

Answer:

(-14,-4)

Step-by-step explanation:

Given the pre-image A=(-7,-2)

If A is dilated with a scale factor of 2

Then the image of A,

A' = (-7,-2) X 2

We multiply each coordinate point by 2

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Therefore, the point A' would be: (-14,-4)

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3 years ago
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mamaluj [8]

\bold{\huge{\red{\underline{Solution}}}}

\bold{\underline{ Given :- }}

  • We have given two figures that rectangle and parallelogram
  • The base and height of parallelogram is 7ft and 5ft
  • The length and breath of rectangle is 3ft and 1ft .

\bold{\underline{ To \: Find :- }}

  • We have to find the area of shaded part?

\bold{\underline{ Let's \: Begin :- }}

Firstly, we will calculate the area of parallelogram then area of rectangle after that we will subtract the area of rectangle from the area of parallelogram then we got the area of shaded part.

\sf{\pink{Area \:of\: parallelogram = Base * height }}

\sf { = 7 × 5 }

\sf { = 35ft }

\sf{ Thus,  \:The \:area\: of \:parallelogram\: is \:35ft}

\sf{\underline{ Now}}

\sf{\blue{Area\: of\: rectangle = L × B }}

\sf{ = 3 × 1 }

\sf{ = 3ft}

\sf{Thus, \:The\: Area\: of \:rectangle\: is\: 3ft}

\sf{\underline{ Now}}

\sf{ Total\: area \:of \:shaded\: part }

\sf{ Area\: of\: parallelogram - Area\: of\: rectangle }

\sf{ = 35 - 3}

\sf{ = 32 ft}

\bold{\pink{Hence, \: Area\: of \:shaded\: part \:is\: 32ft}}

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1. 4

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If 12 boxes of soap weigh 2.4 kg what will be the weight of 8 boxes of the same kind of soap​
RUDIKE [14]

Answer:

1.6kg

explanation:

2.4kg = 12 boxes of the soap

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1.6kg = 8 boxes of the soap

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