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 answer to the problem is 17
The volume would be 201.06
Answer:
(-4, -1/2)
Step-by-step explanation:
to calculate the midpoints bt of the line use the ormula:
Where (x1, x2) is one coordinate point and (y1, y2) is anither coordinate point. Any two points will work, but I chose A (-5,-4) and B (-3, 3).
(-4, -1/2)
1. This must be greater than or equal to 0 (there can't be any negatives!)
8 - .5t ≥ 0.
8 ≥ .5t
t ≤ 16
D.
2. 4(2x-1) = 2x + 35
8x - 4 = 2x + 35
6x = 39
x = 6.5
H.
3. c = .8(x-15) = .8x-12
A.
4. m = number of marble cupcakes
s = number of strawberry shortcake cupcakes
6m + 8.5s = 391
m = 1/2s
6(.5s)+8.5s = 391
11.5s = 391
s = 34
m = 17
F.
5. B
6. Use substitution
2x + 3x = 10
5x = 10
x = 2
7. (sorry I'm not sure!)
8. G
9. L = number of balls
B = number of bats
L + B = 100
B = 100 - L
4.5L+20B = 822
4.5L + 20(100-L) = 822
4.5L -20L + 2000 = 822
1178 = 15.5L
L = 76
10. J
