Not sure is this is right but if the chemist wants to make the solution 35% acidic then he can use all of the 30% acid and 5% of the 45% acid.
Answer:
₹ 4000
Step-by-step explanation:
Sarah's salary =₹ 30000
- one-tenth of her salary to an orphanage= 1/10*₹ 30000= ₹ 3000
- one-third of her salary is spent on food= 1/3*₹ 30000= ₹ 10000
- one-fourth of salary on rent and electricity =1/4*₹ 30000= ₹ 7500
- one-twentieth of her salary on the telephone= 1/12*₹ 30000= ₹ 2500
- donated some amount to the Prime Minister's relief fund = x
- She was left with ₹ 3000.
x= ₹ 30000- (₹ 3000+₹ 10000+₹ 7500+₹ 2500+₹ 3000)= ₹ 4000
Answer:
<em>Johnny had </em><em>5</em><em> items and Bobby had </em><em>13</em><em> items.</em>
Step-by-step explanation:
Let us assume that Johnny has x items and Bobby has y items in their list.
Bobby had 3 more than twice as many items on it than Johnny. So,
----------1
Johnny asked for 8 fewer items than bobby. So,
-----------2
Putting the value of x from equation 2 in equation 1, we get




Putting the value of y in equation 2,


So, Johnny had 5 items and Bobby had 13 items.
9514 1404 393
Answer:
(x, y) = (-2, -1) or (2, 1)
Step-by-step explanation:
Substitute for x in the first equation:
(2y)^2 +3(2y)y = 10
10y^2 = 10
y^2 = 1
y = ±1
x = 2y = ±2
Solutions are (x, y) = (-2, -1) or (2, 1).