p=7, I don't know if this is a question or not
Answer:
10.6
Step-by-step explanation:
1. First, we need to find which digits √(113) is between, and that would be 10 and 11.
2. Okay, so √(113) is between 10 and 11, meaning 113 is between 100 and 121.
3. Since 113 is a little over half between 100 and 121, let's multiply 10.6 by 10.6 to see how close we are to 113:
4. 112.36 is pretty close to 113, therefore, the dot on the number-line should be placed at 10.6.
g) c=1.98w is your answer
w = weight = 5.5
c = cost = 10.89
10.89 = 1.98(5.5)
10.89 = 10.89 (true)
hope this helps
Least common multiple: factor them, then see what they have in common and what is leftover and multiply those expressions:
(x - 2)(x + 3) 10(x + 3)(x + 3)
Common: (x + 3)
Leftover: (x - 2), (10), (x + 3)
Common · Leftover is: (x + 3) · (x - 2) · (10) · (x + 3) = 10(x - 2)(x + 3)²
Answer: LCM is 10(x - 2)(x + 3)²
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
