Answer:
part A: expression 1: 6(8m+2)
expression 2: 6m+12+42m
Part B: 6(m+2+7m)= 6(8m+2)
combine like terms is 6(m+2+7m), so it is 6(2+8m) or 6(8m+2)
6(8m+2)= 6(8m+2)
Part C: 6m+12+42m=6(m+2+7m)
m=0
6(0+2+7(0))=6(0)+12+42(0)
6(2)= 12
12=12
<h2><u>
Answer With Explanation:</u></h2>
<u>Firstly, let's start with <XOZ: =55°</u>
We know that <ZOQ is 70° and angles on a line add up to 180° so we do 180-70=110 110 divided by 2 = 55 so the 2 angles (XOZ & XOP are 55)
<u>Secondly, <OMN, <MON & <ONM = All are 60°</u>
These 2 angles are joined to create an equilateral triangle which always adds up to 180°
So, there are 3 points to this triangle, therefore we divide 180 by 3 which is 60. The angles are 60°
<u>Thirdly, <QON: =55°</u>
This angle lies on the line XON which needs to add up to 180°
As we worked out before, <XOZ was 55°
So, <ZOQ was already given as 70°
We then do 55+70=125 then 180-125=55°
<QON is 55°
(I'm only in Grade 9 LOL)
The ten thousandth number in 14,620 is 1
the number to the right of 1 is 4
4>5
The answer is 10,000
Answer:
Step-by-step explanation:
The wording on this is not the best. It sounds like the 1 zero has even multiplicity (that's because of where the modifier is). On top of that it has an odd power. You could try this. y =x*(x^2+1)^2
The problem is not with the power. It gives x^5. The problem is with the multiplicity of the one place where it crosses. (X^2 + 1) does factor, but it gives a complex root. I'm not sure that's allowed. However, it is the best I can do.
<span>if v belongs to V, then we can find scalars a1,a2,...,an, such that
v=a1*v1+a2*v2+...+an*vn,
L1(v)=L1(a1*v1+a2*v2+...+an*vn)
=a1*L1(v1)+a2*L1(v2)+...+an*L1(vn)
=a1*L2(v1)+a2*L2(v2)+...+an*L2(vn)
=L2(a1*v1+a2*v2+...+an*vn)
=L2(v)</span>
