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2 years ago

What should be done first when evaluating this expression?

Mathematics

1 answer:

gizmo_the_mogwai [7]2 years ago

4 0

Pemdas
You would multiply first

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A plane flying horizontally at an altitude of 1 mi and a speed of 470 mi/h passes directly over a radar station. Find the rate a

kiruha [24]

Answer:

407 mi/h

Step-by-step explanation:

Given:

Speed of plane (s) = 470 mi/h

Height of plane above radar station (h) = 1 mi

Let the distance of plane from station be 'D' at any time 't' and let 'x' be the horizontal distance traveled in time 't' by the plane.

Consider a right angled triangle representing the above scenario.

We can see that, the height 'h' remains fixed as the plane is flying horizontally.

Speed of the plane is nothing but the rate of change of horizontal distance of plane. So, $s=\frac{dx}{dt}=470\ mi/h$

Now, applying Pythagoras theorem to the triangle, we have:

$D^2=h^2+x^2\\\\D^2=1+x^2$

Differentiating with respect to time 't', we get:

$2D\frac{dD}{dt}=0+2x\frac{dx}{dt}\\\\\frac{dD}{dt}=\frac{x}{D}(s)$

Now, when the plane is 2 miles away from radar station, then D = 2 mi

Also, the horizontal distance traveled can be calculated using the value of 'D' in equation (1). This gives,

$2^2=1+x^2\\\\x^2=4-1\\\\x=\sqrt3=1.732\ mi$

Now, plug in all the given values and solve for $\frac{dD}{dt}$ . This gives,

$\frac{dD}{dt}=\frac{1.732\times 470}{2}=407.02\approx 407\ mi/h$

Therefore, the distance from the plane to the station is increasing at a rate of 407 mi/h.

6 0

3 years ago

The month of July has 31 days. What is the greatest

Klio2033 [76]

4 is the most
in the month

3 0

3 years ago

In 16 years, Ben will be 3 times as old as he is right now.

Luda [366]

Answer:

he is 8

Step-by-step explanation:

3 0

3 years ago

I don't know this please help

Komok [63]

Answer:

2/1 I believe, I suck at slope tho, I'm sorry if it's wrong

4 0

2 years ago

Read 2 more answers

What is the average rate of change of the function

Reptile [31]

Answer:

Average rate of change $=-35.7084$

Step-by-step explanation:

Given function is $f(x)=480(0.3)^x$ and we need to find average rate of change of the function from $x=1\ to\ x=5$ .

Average rate of change $=\frac{f(b)-f(a)}{b-a}$

So,

$here\ b=5\ and\ a=1\\f(5)=480(0.3)^5\\=480\times0.00243=1.1664\\and\\f(1)=480(0.3)^1\\=480\times0.3=144$

Average rate of change

$=\frac{f(b)-f(a)}{b-a}\\\\=\frac{f(5)-f(1)}{5-1}\\\\=\frac{1.1664-144}{5-1}\\\\=\frac{-142.8336}{4}= -35.7084$

Hence, average rate of change of the function $f(x)=480(0.3)^x$ over the intervel $x=1\ to\ x=5$ is $=-35.7084$ .

5 0

3 years ago