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.
The answers are
1: False
2: True
3: True
4: False
9514 1404 393
Answer:
r = ±√47
Step-by-step explanation:
-7r^2 = -329 . . . . given
r^2 = 47 . . . . . divide both sides by -7 (multiplication property of equality)
r = ±√47 . . . . square root both sides (square root property of equality)
Q cuts the diagonal PA into 2 equal halves, since the diagonals of rhombus meet at right angles.
<u>Step-by-step explanation:</u>
As given by the statement in the problem,
Q may be the middle point, which cut the diagonal PA into 2 equal halves.
In rhombus, diagonals meet at right angles.
which means that PQ = QA
x+2 = 3x - 14
Grouping the terms, we will get,
3x -x = 14+ 2
2x = 16
dividing by 2 on both sides, we will get,
x = 16/2 = 8
8+2 = 3(8) - 14 = 10 = PQ or QA
