Answer: y=4,728,779,608,739,020,800
Step-by-step explanation: based on the values you have for m & b it would be...
y=4,200 x
y=4,200 x 1,125,899,906,842,624
y=4,728,779,608,739,020,800
Would anyone care to explain how to get through this problem in the photo? I'm at a complete dead end! Any help would be appreci
ated!
1 answer:
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You can get all the terms on one side, and then solve for x by any means:
Factor out a -2:
Factor this equation:
I will use the AC method. To use it, first multiply a and c (in ax^2 + bx +c):
Now, look for two numbers that multiply to -12 and add to -11. Obviously these numbers are -12 and 1. (-12*1 = -12 and -12+1 = -11). Now, because of rules, you set it up in a Punnett square:
4x^2 -12x
x -3
Now, we find common factors of the terms in rows:
_x_______-3__
4x| 4x^2 -12x
1 | x -3
So, you can use this to write an equivalent expression to the quadratic given:
Now, we known the factors are (solve for x): 3 and -1/4.
Factor out a -2:
Factor this equation:
I will use the AC method. To use it, first multiply a and c (in ax^2 + bx +c):
Now, look for two numbers that multiply to -12 and add to -11. Obviously these numbers are -12 and 1. (-12*1 = -12 and -12+1 = -11). Now, because of rules, you set it up in a Punnett square:
4x^2 -12x
x -3
Now, we find common factors of the terms in rows:
_x_______-3__
4x| 4x^2 -12x
1 | x -3
So, you can use this to write an equivalent expression to the quadratic given:
Now, we known the factors are (solve for x): 3 and -1/4.
Answer:
The viscosity at 140°C is predicted to be 7.2
Step-by-step explanation:
The function that model the relationship between viscosity and temperature = Quadratic model
The general form of a quadratic equation is y = a·x² + b·x + c
Therefore, we have;
When y = 10.8, x = 110, which gives;
10.8 = a·110² + b·110 + c = 12100·a + 110·b + c
10.8 = 12100·a + 110·b + c
When y =8.2, x = 130
8.2 = a· 130² + b· 130 + c = 16900·a + 130·b + c
8.2 = 16900·a + 130·b + c
When y = 160, x = 5.8
5.8 = a·160² + b·160 + c = 25600·a + 160·b + c
5.8 = 25600·a + 160·b + c
The three equations above can be listed as follows;
10.8 = 12100·a + 110·b + c
8.2 = 16900·a + 130·b + c
5.8 = 25600·a + 160·b + c
Solving using matrices gives;
From which we have;
a = 0.001, b = -0.37, c = 39.4
Substituting gives;
y = 0.001·x² - 0.37·x + 39.4
When x = 140
y = 0.962·140² - 0.37·140 + 39.4= 7.2
The viscosity at 140°C = 7.2.