Answer:
x=±√−9
No real solutions.
Step-by-step explanation:
Step 1: Add 3x^2 to both sides.
27+3x2=−3x2+3x2
3x2+27=0
Step 2: Subtract 27 from both sides.
3x2+27−27=0−27
3x2=−27
Step 3: Divide both sides by 3.
3x2
/3
=
−27
/3
x2=−9
Step 4: Take square root.
x=±√−9
Her is the data of major river system in India on the basis of their pollution level:
Sabarmati :
Around 4 mg/L
On the basis of the statistics taken in 2010 are:
Markanda: 590 mg Oxygen /liter
Amlakhadi: 353 mg Oxygen /liter
Kali: 364 mg Oxygen /liter
Yamuna canal: 247 mg Oxygen /liter
The Yamuna near Delhi: 70 mg Oxygen /liter
Betwa: 58 mg Oxygen /liter
The levels of 8 mg/L or less is acceptable and below 2 mg/L is also considered as a good water.
Ganga :
At Rishikesh: 1.5 mg Oxygen/L.
At Kanpur: 5 mg/L
Brahmaputra:
At Gauhati - very high.
At other locations: 2 mg /L or less
Tapi:
In certain areas: 2 mg/L .
In cities and towns: 13 mg/L
Narmada:
Less than 2.5 mg/L
Mahanadi:
Less than 1 mg Oxygen / Litre to 4 mg Oxygen / Litre in cities or Chattisgarh
Krishna:
1 mg Oxygen / Litre to 3 mg Oxygen / Litre 10 mg Oxygen / Litre in certain cities/towns.
Pune:
Around 28 mg/L
Cauvery :
From 1 mg/L to 3 mg/L
Ganga :
From 4 mg/l to 8 mg/l
Godavari :
Around 5 mg/l
Vitharini (Bitarni) :
Around 2 mg/l
Narmada :
Around 2 mg/l
Mahi :
Around 2.5 mg//l
We have the following:
AC=root(a^2 + b^2)
Then,
BD=root((a-0)^2 + (0-b)^2)
BD=root((a)^2 + (-b)^2)
BD=root((a)^2 + (b)^2)
therefore,
AC=AB
Answer:
AC=AB
Answer:
5
Step-by-step explanation:
2x+3(x+4)=37
2(5)+3(5+4)=37
10+3(9)=37
10+27=37
37=37
