Is the system of equations consistent and independent, consistent and dependent, or inconsistent? y=−3x 12y=−6x 2 Select the cor
1 answer:
System of equations helps to compare two real-life situations. The system is an Independent Consistent System.
<h3>What is a System of the equation?</h3>
Inconsistent System
A system of equations to have no real solution, the lines of the equations must be parallel to each other.
Consistent System
1. Dependent Consistent System
A system of the equation to be Dependent Consistent System the system must have multiple solutions for which the lines of the equation must be coinciding.
2. Independent Consistent System
A system of the equation to be Independent Consistent System the system must have one unique solution for which the lines of the equation must intersect at a particular.
Given to us
y = −3x + 12
y = −6x + 2
As we have the value of y in both equations equate it against each other,
We know the value of x, substitute the value of x in the equation of y,
As the value of both x and y is unique, therefore, there is a unique solution for the given system of equations.
Thus, the system has a unique solution, therefore, the system is an Independent Consistent System.
Learn more about System of equation:
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Answer:
Option C is correct.
Bobby has $300 in the yearbook fund and earns $13.50 for each yearbook sold.
Step-by-step explanation:
Slope intercept form: An equation of line is given by: ......[1] where m is the slope and b is the y-intercept
Given the linear equation:
Here, x represents the yearbook sold.
On comparing with the equation [1] we have;
m = $13.50
b =300
therefore, the best scenario best matches the linear relationship is:
Bobby has $300 in the yearbook fund and earns $13.50 for each yearbook sold.
Answer:
7. $8123.79
8. 0.012 g
Step-by-step explanation:
It often pays to follow directions. The attachment shows the use of a TI-84 graphing calculator to find the answers.
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You will notice that the answer to problem 8 does not agree with any of the offered choices. The time period of 22.8 years is 12 times the half-life of the substance, so there will be (1/2)^12 = 1/4096 of the original amount remaining. The time periods corresponding to the amounts shown range from 1.37 years to 16.4 years.
For half-life problems, I find it convenient to use the decay factor (0.5^(1/half-life)) directly, rather than convert it to e^-k. If you do convert it to the form ...
e^(-kt)
the value of k is (ln(2)/half-life), about 0.3648143056.
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For multiple choice problems where the choices make no sense, I like to suggest you ask your teacher to show you how to work the problem. (Alternatively, use the "Report this question" or "Ask a tutor" button sometimes provided.)
