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2 years ago

For every quad, there are 4 wheels. Complete the

Mathematics

1 answer:

PIT_PIT [208]2 years ago

7 0

The one next to 3 is 12, the one next to 7 is 28, the one next to 12 is 48. All you have to do is multiply the quads by four and you get the number of wheels.

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What is the median of the data displayed in this box-and-whisker plot? A.41 B.49 C.55 D.58

lutik1710 [3]

The median in a whisker plot is the dot or line in the box on the line. In this case, the dot is at 41, so the median is 41.

ANSWER: A) 41 is the median

Hope this helps! :)

7 0

3 years ago

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18 is what percent of 60

Arisa [49]

<span>first you set up a proportion 18 over ? = 60 over 100 then cross multiply. 18*100=1800 then divide that by 60 and you get 30 :)</span>

5 0

3 years ago

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Help!! Please!!!!!!!!!!!!!!

Liono4ka [1.6K]

Answer:

120°

Step-by-step explanation:

All triangles are equal to 180° so you would subtract the total (180) from the given

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2 years ago

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According to Lambert's law, the intensity of light from a single source on a flat surface at point P is given by Upper L equal

malfutka [58]

Answer:

(a) $L = k*(1 - sin^{2}(\theta))$

(b) L reaches its maximum value when θ = 0 because cos²(0) = 1

Step-by-step explanation:

Lambert's Law is given by:

$L = k*cos^{2}(\theta)$ (1)

(a) We can rewrite the above equation in terms of sine function using the following trigonometric identity:

$cos^{2}(\theta) + sin^{2}(\theta) = 1$

$cos^{2}(\theta) = 1 - sin^{2}(\theta)$ (2)

By entering equation (2) into equation (1) we have the equation in terms of the sine function:

$L = k*(1 - sin^{2}(\theta))$

(b) When θ = 0, we have:

$L = k*cos^{2}(\theta) = k*cos^{2}(0) = k$

We know that cos(θ) is a trigonometric function, between 1 and -1 and reaches its maximun values at nπ, when n = 0,1,2,3...

Hence, L reaches its maximum value when θ = 0 because cos²(0) = 1.

I hope it helps you!

5 0

2 years ago

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At 8:00 am, here's what we know about two airplanes: Airplane #1 has an elevation of 80870 ft and is decreasing at the rate of 4

wel

Let's begin by listing out the information given to us:

8 am

airplane #1: x = 80870 ft, v = -450 ft/ min

airplane #2: x = 5000 ft, v = 900ft/min

We must note that the airplanes are moving at a constant speed. The equation for the airplanes is given by:

$\begin{gathered} E=x_1+vt----1 \\ E=x_2+vt----2 \\ where\colon E=elevation,ft;x=InitialElevation,ft; \\ v=velocity,ft\text{/}min;t=time,min \\ x_1=80,870ft,v=-450ft\text{/}min \\ E=80870-450t----1 \\ x_2=5,000ft,v=900ft\text{/}min \\ E=5000+900t----2 \end{gathered}$

We equate equations 1 & 2 to get the time both airlanes will be at the same elevation. We have:

$\begin{gathered} 5000+900t=80870-450t \\ \text{Add 450t to both sides, we have:} \\ 900t+450t+5000=80870-450t+450t \\ 1350t+5000=80870 \\ \text{Subtract 5000 from both sides, we have:} \\ 1350t+5000-5000=80870-5000 \\ 1350t=75870 \\ \text{Divide both sides by 1350, we have:} \\ \frac{1350t}{1350}=\frac{75870}{1350} \\ t=56.2min \\ \\ \text{After }56.2\text{ minutes, both airplanes will be at the same elevation} \end{gathered}$

The elevation at that time (when the elevations of the two airplanes are the same) is given by substituting the value of time into equations 1 & 2. We have:

$\begin{gathered} E_1=80870-450t \\ E_1=80870-450(56.2) \\ E_1=80870-25290 \\ E_1=55580ft \\ \\ E_2=5000+900t \\ E_2=5000+900(56.2) \\ E_2=5000+50580 \\ E_2=55580ft \\ \\ \therefore E_1\equiv E_2=55580ft \end{gathered}$

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6 months ago