10 L of 30 % saline solution can be formed by mixing 4 L of 60 % saline solution and 6 L of 10 % saline solution.
Step-by-step explanation:
Let x be the number of liters of 60% saline solution
Now we require 10 L of 30% saline solution.
Liter soln % liters saline %
30 % 10 0.3
60 % x 0.6
10 % 10-x 0.1
Now forming the algebraic equation,
0.6x + 0.1 (10-x) = 10 (0.3)
0.6x + 1 - 0.1 x = 3
0.5 x = 2
x = 4 ( 4 l of 60 % solution is required. So 10 % saline solution required is 10 - 4 = 6 L).
Hence, 10 L of 30 % saline solution can be formed by mixing 4 L of 60 % saline solution and 6 L of 10 % saline solution.
Answer:
Ihorangi uses 7.65 litres of petrol to and from work
Step-by-step explanation:
If the car Ihorangi drives consumes 7.5 L/100 km
Then for 1 km journey, the car will consume 7.5/100 L
= 0.075 L
If Ihorangi travels 51 km to work and 51 km from work
Therefore Ihorangi travel (51 + 51) km daily to and from work
= 102 km daily
The fuel he consumes daily is 102*0.075 L
= 7.65 L
Answer:
Please follow me back
Step-by-step explanation:
The Answer is 158
Answer:
f(2) = 12
f(x) = 7, x = -3, 1
Step-by-step explanation:
<u>a)</u>
plug in x as 2
f(x) = 2^2 + 2(2) + 4
f(x) = 4 + 4 + 4
f(x) = 12
<u>b)</u>
replace f(x) with 7
7 = x^2 + 2x + 4
x^2 + 2x - 3 (move 7 to other side)
Factor
ac: -3x^2
b: 2x
split b into 3x, -x
(x^2 -x) + (3x - 3)
↓ ↓
x(x-1) + 3(x-1)
Factor: (x-1)(x+3) = 0
Solve using Zero Product Property:
x - 1 = 0, x + 3 = 0
x = 1, x = -3
The answer is -4,0 4,8 because u would solve for one variable in one of the equations and then substitute the result into the other
