Next, solve as you would solve any two-step inequality. Since 2 is added to , you can get by itself on one side of the inequality by subtracting 2 from both sides. Then, to get by itself on one side of the inequality, you need to multiply both sides by 2.D

8 0

4 years ago

Simplify 2a^2b^3 (4a^2 + 3ab^2 - ab) =?

stellarik [79]

The correct answer is D.D. 8a^4b^3 + 6a^3b^5 - 2a^3b^4

You can tell this by the lead term alone. When you mutliply the term outside to the lead term, D is the only one that has that term.

2a^2b^3*4a^2 = 8a^4b^3

D. 8a^4b^3 + 6a^3b^5 - 2a^3b^4

7 0

4 years ago

What types of angles are these? see attachment

Alona [7]

Answer:

I believe its corresponding although they dont seem like they're the same amount of degrees but if it isnt corresponding, everything else doesnt suit it

Hope i helped :)

3 0

3 years ago

Solve for X

sweet-ann [11.9K]

Answer:

x = 1035 = MXXXV

Step-by-step explanation:

<u>Roman Numbers </u>

It's an ancient number system developed in Rome where numbers are represented by the letters {I,V,X,L,C,D,M}. The value of each are, respectively {1,5,10,50,100,500,1000}. If some letter is repeated, then it's value must be multiplied by the number of times the letter is present. For example, XXVII has two X's and two I's, so the conversion is 10+10+5+1+1=27.

If a less-valued letter appears before a greater one, it subtracts. For example CDI=-100+500+1=401

Our numbers are

MMMMMMMMCDXV=8*1000-100+500+10+5=8415

MDCCCLXIII = 1000+500+300+50+10+3=1863

DLXI = 500+50+10+1=561

XXVII = 20+5+2=27

We now operate in our numerical system

$\displaystyle x = \frac{MMMMMMMMCDXV * MDCCCLXIII}{DLXI * XXVII}$