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3 years ago

Scientific notation of 47300

1 answer:

6 0

4.7 x 10^4
or 5 x 10^4

<span>The scientific notation of 70, 030, 000. To find the scientific notation of this value </span><span><span>
1.  </span>We first move the period which separates the whole number from the decimal number which is located after the numbers of the given value.</span>
<span><span>2.  </span>We move it in the very recent order number which is seventy million, seven and zero.</span> <span><span>
3.  </span>It becomes 7.003</span>
<span><span>4.  </span>Thus we count how many moves we did from the tens to the ten million order place.</span> <span><span>
5.  </span>7.003 x 10^7 </span>

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3 years ago

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4 years ago

Read 2 more answers

Pls solve part b) iii thanks

Viktor [21]

The bearing of the tree from Q is 296.565°

<h3>How to determine the height of the tree?</h3>

The figure that illustrates the bearing and the distance is added as an attachment

The given parameters are:

Base of the tree, b = 50 meters

Angle (x) = 32 degrees

Calculate the height (h) of the tree using:

tan(x) = height/base

So, we have:

tan(32°) = h/50

Make h the subject

h= 50 × tan(32°)

Evaluate

h = 31.24

Hence, the height of the tree is 31.24 meters

<h3>How to determine the distance between Q and the base of the tree?</h3>

The distance (d) between Q and the base of the tree

This is calculated using the following Pythagoras theorem

d = √(100² + 50²)

Evaluate

d = 111.80

Hence, the distance between Q and the base of the tree is 111.80 meters

<h3>How to determine the angle of elevation?</h3>

The angle of elevation (x) using the following tangent trigonometric ratio

tan(x) = h/d

This gives

tan(x) = 31.24/111.80

Evaluate the quotient

tan(x) = 0.2794

Take the arc tan of both sides

x = 15.61

<h3>The bearing of the tree from Q </h3>

This is calculated using:

Angle of bearing = 270 + arctan(50/100)

Evaluate the arc tan

Angle of bearing = 270 + 26.565

Evaluate the sum

Angle of bearing = 296.565

Hence, the bearing of the tree from Q is 296.565 degrees