Answer: tea = 15 rupees per kg
sugar= 3 rupees per kg
Step-by-step explanation:
Hi, to answer this question we have to write a system of equations with the information given:
<em>"Two kg of tea and 3 kg of sugar cost rupees 39 in january 1997":
</em>
2 t + 3 s =39 (a)
Where:
- t= price of 1 kg of tea
- s = price of 1 kg of sugar
<em>"in march 1997 the price of the tea increased by 25% (1.25)and the price of the sugar increased by 20%(1.20) and the same quantity of tea and sugar cost rupees 48.30.
"</em>
2(t1.25)+3(s1.2) = 48.30 (b)
- <em>Solving for t in (b)
</em>
2t =39-3s
t = (39 -3s)/2
t = 19.5-1.5s
- <em>Replacing the value of t in (b)
</em>
2 x ((19.5-1.5s)1.25)+ 3 ( 1.2s) =48.30
2x ( 24.375 -1.875s) +3.6s =48.30
48.75 -3.75s+3.6s= 48.30
48.75-48.30 = 3.75s-3.6s
0.45= 0.15s
0.45/0.15 =s
3 =s
- <em>Replacing the value of s in (a)
</em>
2 t + 3 (3) =39
2 t + 9 =39
2 t =39 -9
2 t =30
t = 30/2
t= 15
Prices in january:
tea = 15 rupees per kg
sugar= 3 rupees per kg
Feel free to ask for more if needed or if you did not understand something.
Answer:
Step-by-step explanation
6n - 6 = 2 (n+1)
Simplify 2 (n+1)
6n - 6= 2n +2
Move all terms related to n to the left side of the equation
4n -6 =2
Move all terms not related to n to the right side of the equation
4n =8
Divide each term by 4 then simplify
n=2
1. You need to multiply the denominator by something that will make the content of the radical be a square—so that when you take the square root, you get something rational. Easiest and best is to multiply by √6. Of course, you must multiply the numerator by the same thing. Then simplify.
2. Identify the squares under the radical and remove them.
Answer:
$100 + $10<em>m</em>
Step-by-step explanation:
A = Account Balance = 100
After M months
A = $100 + $10m
