Answer:
y^3 + 5y + 8
Step-by-step explanation:
Collect like terms
4y^3 - 3y^3 + 2y + 2y + y + 4 + 4
= 1y^3 + 5y + 8 OR y^3 + 5y + 8
Answer:
Cost of adult ticket = $12.5
Cost of child ticket = $7.5
Step-by-step explanation:
Given:
Cost of 6 adult ticket and 5 child ticket = $112.5
Cost of 8 adult ticket and 4 child ticket = $130
Find:
Equation and solution
Computation:
Assume;
Cost of adult ticket = a
Cost of child ticket = b
So,
6a + 5b = 112.5....eq1
8a + 4b = 130 ......eq2
Eq2 x 1.25
10a + 5b = 162.5 .....eq3
eq3 - eq1
4a = 50
Cost of adult ticket = $12.5
8a + 4b = 130
8(12.5) + 4b = 130
Cost of child ticket = $7.5
Step-by-step explanation:
if there are no typos, then the second and the fourth answers are correct.
1 - 1.7 = -0.7 or -2×0.35
-2x = -2(x)
Could you type the answer into this answers comments? because I think I see 8452 + something but i cant see what that is
In order to have infinitely many solutions with linear equations/functions, the two equations have to be the same;
In accordance, we can say:
(2p + 7q)x = 4x [1]
(p + 8q)y = 5y [2]
2q - p + 1 = 2 [3]
All we have to do is choose two equations and solve them simultaneously (The simplest ones for what I'm doing and hence the ones I'm going to use are [3] and [2]):
Rearrange in terms of p:
p + 8q = 5 [2]
p = 5 - 8q [2]
p + 2 = 2q + 1 [3]
p = 2q - 1 [3]
Now equate rearranged [2] and [3] and solve for q:
5 - 8q = 2q - 1
10q = 6
q = 6/10 = 3/5 = 0.6
Now, substitute q-value into rearranges equations [2] or [3] to get p:
p = 2(3/5) - 1
p = 6/5 - 1
p = 1/5 = 0.2
