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2 years ago

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Six cell phones weigh 46.2 ounces. Fourteen cell phones weigh 107.8 ounces. This relationship is an example of a proportional re

lationship. What is the constant of proportionality

1 answer:

Alika [10]2 years ago

5 0

Answer:

The constant is 7.7 . This is because the quantities are directly proportional so constant of proportionality is a/b or 46.2/6

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Find the area of this parallelogram, top= 14 left= 11 right = 10

Kamila [148]

Answer:

140 cm^2

Step-by-step explanation:

Formula for area of a Parallelogram: base*height

14 cm * 10cm = $140 cm^{2}$

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2 years ago

Find g (x) - f (x). <br><br> g (x) = x^2 + 1 and f (x) = 2x + 5

allochka39001 [22]

Answer:

x^2 - 2x - 4

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2 years ago

Anna would like to purchase a new bike that costs $205. She already has $31 saved. If she saves a maximum of $12 a week, the fol

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Answer:

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3 years ago

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Does it seem plausible that employment has a normal distribution for each gender

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yes it is reasonable that employment has a normal distribution for each gender

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3 years ago

You are looking at the New York ball drop on New Year’s Eve at a distance of 100 m away from the base of the structure. If the b

Fofino [41]

The question is an illustration of related rates.

The rate of change between you and the ball is 0.01 rad per second

I added an attachment to illustrate the given parameters.

The representations on the attachment are:

$\mathbf{x = 100\ m}$

$\mathbf{\frac{dy}{dt} = 2\ ms^{-1}}$ ---- the rate

$\mathbf{\theta = \frac{\pi}{4}}$

First, we calculate the vertical distance (y) using tangent ratio

$\mathbf{\tan(\theta) = \frac{y}{x}}$

Substitute 100 for x

$\mathbf{y = 100\tan(\theta) }$

$\mathbf{\tan(\theta) = \frac{y}{100}}$

Differentiate both sides with respect to time (t)

$\mathbf{ \sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot \frac{dy}{dt}}$

Substitute values for the rates and $\mathbf{\theta }$

$\mathbf{ \sec^2(\pi/4) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

This gives

$\mathbf{ (\sqrt 2)^2 \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

$\mathbf{ 2 \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

Divide both sides by 2

$\mathbf{ \frac{d\theta}{dt} = \frac{1}{100} }$

$\mathbf{ \frac{d\theta}{dt} = 0.01 }$

Hence, the rate of change between you and the ball is 0.01 rad per second

Read more about related rates at:

brainly.com/question/16981791

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2 years ago