Answer:
Question 1: 4.5(10^−7)(2(10^4))
=
9/1000
(Decimal: 0.009)
Question 2:
mass of neutron/mass of electron = 2*10-24/(9*10-28)
2x10^(-24)/9x10^(-28)
d is closest.
Question 3:
4(10^3)(12(10^5))
=4000*1200000
=4*1000*12*100000
=(4*12)*(1000*100000)
=48*100000000
=4800000000
And these things : ^ mean raising the number to become exponets and when I put the number into a bold text thats the answer!
<em>Hope</em><em> </em><em>this</em><em> </em><em>helps</em><em>!</em><em> </em><em>if</em><em> </em><em>so</em><em> </em><em>pls</em><em> </em><em>mark</em><em> </em><em>brainliest</em><em> </em><em>and</em><em> </em><em>heart</em><em>/</em><em>rate</em><em>!</em>
Answer:
y+4=–109(x-7)
please mark as brainliest
Answer:
1.) BC, AD, FI.
2.) Parallel.
Step-by-step explanation:
I can't get the other two I'm sorry mate like I really gotta go. Before my mum gets mad. Got chores. Good luck.
Do six multiply fifteen which equals to 90
Answer:
The area of the shaded region is 42.50 cm².
Step-by-step explanation:
Consider the figure below.
The radius of the circle is, <em>r</em> = 5 cm.
The sides of the rectangle are:
<em>l</em> = 11 cm
<em>b</em> = 11 cm.
Compute the area of the shaded region as follows:
Area of the shaded region = Area of rectangle - Area of circle
![=[l\times b]-[\pi r^{2}]\\\\=[11\times 11]-[3.14\times 5\times 5]\\\\=121-78.50\\\\=42.50](https://tex.z-dn.net/?f=%3D%5Bl%5Ctimes%20b%5D-%5B%5Cpi%20r%5E%7B2%7D%5D%5C%5C%5C%5C%3D%5B11%5Ctimes%2011%5D-%5B3.14%5Ctimes%205%5Ctimes%205%5D%5C%5C%5C%5C%3D121-78.50%5C%5C%5C%5C%3D42.50)
Thus, the area of the shaded region is 42.50 cm².
