Student 1 and student 3 are equivalent
See below for the terms, coefficients, and constants in the variable expressions
<h3>How to determine the terms, coefficients, and constants in the variable expressions?</h3>
To determine the terms, coefficients, and constants, we use the following instance:
ax + by + c
Where the variables are x and y
- Then the terms are ax, by and c
- The coefficients are a and b
- The constant is c
Using the above as guide, we have:
A) 2b + 2ac+5
- Terms: 2b, 2ac, 5
- Coefficient: 2, 2 and 5
- Constant 5
B) 34abx + 16y +1
- Terms: 34abx, 16y, 1
- Coefficient: 34ab, 16
- Constant: 1
C) st +4u + v
- Terms: st, 4u, v
- Coefficient: 4
D) 14xy + 6
- Terms: 14xy, 6
- Coefficient: 14, 6
- Constant 6
E) 14x + 12y
- Terms: 14x, 12y
- Coefficient: 14, 12
F) 3+ 6-7+a
- Terms: 3, 6, -7, a
- Coefficient: 1
- Constant: 3, 6, -7
Read more about terms, coefficients, and constants at:
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Answer:
Ã=7/6
Step-by-step explanation:
<em>2/3(x -6) -y </em>
<h3><em>put the value of x= 9 and y=5/6 </em></h3>
=2/3(9-6)-5/6
=2/3×3-5/6
=2/1-5/6
=12-5/6
=7/6
<em>hey mate hope it's help you</em>......
Answer: this is the right answer Graph A represents j(x). The y-intercepts for f(x) and g(x) can be 1 and 3. And The rate of change of the sum of f(x) and g(x) is greater than that of either function. your welcome
Step-by-step explanation:
Answer:
The probably genotype of individual #4 if 'Aa' and individual #6 is 'aa'.
Step-by-step explanation:
In a non sex-linked, dominant trait where both parents carry and show the trait and produce children that both have and don't have the trait, they would each have a genotype of 'Aa' which would produce a likelihood of 75% of children that carry the dominant traint and 25% that don't. Since the child of #1 and #2, #5, does not exhibit the trait, nor does the significant other (#6), then they both must have the 'aa' genotype. However, since #4 displays the dominant trait received from the parents, it is more likely they would have the 'Aa' genotype as by the punnet square of 'Aa' x 'Aa', 50% of their children would have the 'Aa' phenotype.
