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
The factors of given polynomial are:![(2x+9)(3x+7)](https://tex.z-dn.net/?f=%282x%2B9%29%283x%2B7%29)
Further explanation:
Factorization is either done by using quardatic formula or directly making factor.
For making factors, the constant term is multiplied with the coefficient of x^2 and pairs of factors are made such that their sum or difference is equal to the co-efficient of x.
Given
![6x^2+41x+63](https://tex.z-dn.net/?f=6x%5E2%2B41x%2B63)
63x6 gives us: 378
Factor pairs of 378 are:
182,2
126,3
54,7
42,9
14,27 => 41
So the pair we are going to use is 14,27
![6x^2+41x+63\\=6x^2+14x+27x+63\\=2x(3x+7)+9(3x+7)\\=(2x+9)(3x+7)\\](https://tex.z-dn.net/?f=6x%5E2%2B41x%2B63%5C%5C%3D6x%5E2%2B14x%2B27x%2B63%5C%5C%3D2x%283x%2B7%29%2B9%283x%2B7%29%5C%5C%3D%282x%2B9%29%283x%2B7%29%5C%5C)
So the factors of given polynomial are:![(2x+9)(3x+7)](https://tex.z-dn.net/?f=%282x%2B9%29%283x%2B7%29)
Keywords: Factorization, polynomials
Learn more about factorization at:
#LearnwithBrainly
Answer:
N = 74(1/2)^(t/2.8)
Step-by-step explanation:
The exponential function expressing a half-life relation can be written ...
amount = (initial amount) × (1/2)^(t/(half-life))
For the numbers given in this problem, this is ...
N = 74(1/2)^(t/2.8)
__
Some folks like to express these relations in the form ...
N = 74e^(-kt)
In this form, the value of k is ...
k = ln(2)/(half-life) ≈ 0.693147/2.8 ≈ 0.24755
N = 74e^(-0.24755t)
Answer:
0.2611
Step-by-step explanation:
Given the following information :
Normal distribution:
Mean (m) length of time per call = 3.5 minutes
Standard deviation (sd) = 0.7 minutes
Probability that length of calls last between 3.5 and 4.0 minutes :
P(3.5 < x < 4):
Find z- score of 3.5:
z = (x - m) / sd
x = 3.5
z = (3.5 - 3.5) / 0.7 = 0
x = 4
z = (4.0 - 3.5) / 0.7 = 0.5 / 0.7 = 0.71
P(3.5 < x < 4) = P( 0 < z < 0.714)
From the z - distribution table :
0 = 0.500
0.71 = very close to 0.7611
(0.7611 - 0.5000) = 0.2611
P(3.5 < x < 4) = P( 0 < z < 0.714) = 0.2611