Answer:
<u>2.256 x 10 to the power of 25</u>
Step-by-step explanation:
There are 3.76 x 10 to the power of 22 atoms in 1 gram of oxygen
So, there are x atoms in 600 grams of oxygen
Using the ratio and proportional
So, the number of atoms = 600 * 3.76 x 10 to the power of 22 =
So, the answer in scientific notation as A.BCD x 10 to the power of E
Will be <u>2.256 x 10 to the power of 25</u>
Step-by-step explanation:
This is a word problem that we must translate to mathematical expression before we can solve.
let the individual saving be the first letters of the names
Jada= J
Lin= L
Elena= E
1. Jada saved $20 less than Lin
J=L-20
2. Lin saved 3 times as much money as Elena
L= 3E
3. Hence Elena saved
E
Together they saved $120.
hence
(L-20)+3E+E= 120
L-20+3E+E= 120
collecting like terms we hav
L+4E=120+20
L+4E= 140
but we know that L= 3E
3E+4E= 140
7E= 140
divide both sides by 7 we have
E= 140/7
E= $20
This means Elena saves $70
also L= 3E
L= 3*20
L= $60
this means that Lean saved $60
also J=L-20
J= 60-20
J= $40
This means that Jada saved $40
<h3>Given</h3>
A regular polygon with area 500 ft² and apothem 10 ft
Cost of fence is $7.95 per ft
<h3>Find</h3>
Part III The cost of fence around an area scaled to 60 times the size
<h3>Solution</h3>
You don't want to think too much about this, because if you do, you find the regular polygon has 3.087 sides. The closest approximation, an equilateral triangle, will have an area of 519.6 ft² for an apothem of 10 ft.
For similar shapes of scale factor "s", the larger shape will have an area of s² times that of the smaller one. Here, it appears the area scale factor s² is 60, so the linear scale factor is
... s² = 60
... s = √60 ≈ 7.7460
The perimeter fence of the 500 ft² area is presumed to be 100 ft long (twice the area of the polygon divided by the apothem—found in Part I), so the perimeter fence of the industrial farm is ...
... (100 ft)×7.7460 = 774.60 ft
and the cost to construct it is
... ($7.95/ft)×(774.60 ft) ≈ $6158
Answer:
The attached image shows the equation graphed on a graph. From there you can plot the points.
