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Anuta_ua [19.1K]
3 years ago
12

Question 1

Mathematics
1 answer:
ra1l [238]3 years ago
3 0

Answer:

UMMMMM I DONT KMOW

Step-by-step explanation:

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What should the IUPAC name for a binary covalent compound lack? prefixes Roman numerals an -ide ending the name of a nonmetal.​
vitfil [10]

The  IUPAC name for a binary covalent compound lack prefixes such as mono, di, tri tetra etc.

A binary compound is a compound that is composed of only two elements. Many binary compounds could be made up of a metal and a nonmetal or even two nonmetals as the case may be. Examples of binary compounds include; SO2, NaH, K2S etc.

We must note that a binary covalent compound uses Roman numerals to indicate the oxidation state of the central atom in the compound. For instance, SO2 is called Sulfur IV oxide.

Hence, the  IUPAC name for a binary covalent compound lack prefixes such as mono, di, tri tetra etc.

Lear more about IUPAC: brainly.com/question/11587934

5 0
3 years ago
PLEASEE ANSWER QUESTION A AND B ASAPPPPPP
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Answer: A. its basically symmetry

B. I don't know

Step-by-step explanation:

6 0
3 years ago
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T=v /6 + 5<br> a) Work out the value of T when v = 42
Leno4ka [110]

The value of T in the equation if v = 42 is 12

<h3>How to solve algebra</h3>

Given:

T= v /6 + 5

If v = 42, find T

  • substitute 42 for v in the equation

T= v /6 + 5

T = 42/6 + 5

  • solve the fraction first

T = 7 + 5

T = 12

Therefore, the value of T in the equation if v = 42 is 12

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brainly.com/question/4344214

8 0
3 years ago
Need help on geometry!!!
aleksandr82 [10.1K]

Answer:

15. a,    16. ∠GOH = 26

Step-by-step explanation:

15. A straight angle looks as a line.

16. ∠GOI = ∠GOH + ∠HOI

47 = ∠GOH + 21

∠GOH = 47 - 21 =26

∠GOH = 26

5 0
4 years ago
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Circle 1, Circle 2, and Circle 3 have the same center and have radii, respectively, of r₁ cm, r₂ cm, r₃ cm, where r₁ &lt; r₂ &lt
ira [324]

Answer:

a) A₁ /A₂  =  r₁² / (r₂²   - r₁²)

b) A₂ /A₃  = (r₂²   - r₁²) / (r₃²   - r₂²)

Step-by-step explanation:

We have Circle 1  and area  A₁  

Area of circle 2 outside circle 1     =  A₂

Area of circle 3 outside circle 2    = A₃

On the other hand we have

A₁  = π*r₁²   area of circle 1

A₂´ =  π*r₂² area of circle 2

A₃´ = π*r₃²  area of circle 3

All areas in cm²

a) A₁ /A₂        

A₁  =  π*r₁²

According to problem statement  A₂  = π*r₂²  -   A₁

A₂  =  π*r₂²  -   π*r₁²      ⇒         A₂  =  π* (r₂²   - r₁²)

Then  A₁ /A₂  =   π*r₁² / [π* (r₂²   - r₁²)]

A₁ /A₂  =  r₁² / (r₂²   - r₁²)

b)  A₂ /A₃        

A₂  =   π* (r₂²   - r₁²)

And

A₃   =  π* (r₃²   - r₂²)

Therefore

A₂ /A₃  =   π* (r₂²   - r₁²) / π* (r₃²   - r₂²)  ⇒  A₂ /A₃  = (r₂²   - r₁²) / (r₃²   - r₂²)

A₂ /A₃  = (r₂²   - r₁²) / (r₃²   - r₂²)

6 0
3 years ago
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