Answer:
y = 130
Step-by-step explanation:
(4x+8) + (3x-19) + (x+7) = 180
4x + 8 + 3x - 19 + x + 7 = 180
(4x + 3x + x) + (8 -19 + 7) = 180
8x - 4 = 180
8x -4 +4 = 180 +4
8x = 184
8x/8 = 184/8
x = 23
3(23) - 19 = 50
50 + y = 180
180 - 50 = y
y = 130
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
Answer:
BC = 1.71
Step-by-step explanation:
well to start we have to know the relationship between angles, legs and the hypotenuse in a right triangle
α = 70°
a: adjacent = BC
h: hypotenuse = 5
sin α = o/h
cos α= a/h
tan α = o/a
we see that it has (angle, adjacent, hypotenuse)
we look at which meets those data between the sine, cosine and tangent
is the cosine
cos α = a/h
Now we replace the values and solve
cos 70 = a/5
0.34202 = a/5
0.34202 * 5 = a
1.7101 = a
round to the neares hundredth
a = 1.7101 = 1.71
BC = 1.71
