Brianna's thinking is incorrect.
Expessions A, and C. are equivalent.
For C, you add 5x and x together (because they have the same variable) to get 6x - 4.
For A, you subtract 9x and 3x (because they have the same variable) to get 6x- 4.
Answer:
The answer is C.
Step-by-step explanation:
Just go to Desmos.com and plug it in.
Answer:
The 17th term in arithmetic sequence is 68
Step-by-step explanation:
The general formula of arithmetic sequence is:
aₙ = a₁ + (n – 1)d.
We are given a₆ = 101 and a₉ = 83 and we need to find a₁₇
To find the term a₁₇ we should know a₁ and d. So we would find both
a₆ = a₁ +(6-1)d
101 = a₁ +(5)d
101 = a₁ +5d eq(1)
and
a₉ = a₁ +(9-1)d
83 = a₁ + 8d eq(2)
Subtracting eq(2) from eq(1)
101 = a₁ +5d
83 = a₁ + 8d
- - -
__________
18 = -3d
=> d = 18/-3
=> d = -6
Putting value of d in eq(1)
101 = a₁ + 5d
101 = a₁ + 5(-3)
101 = a₁ -15
=> a₁ = 101+15
=> a₁ = 116
Now finding a₁₇:
aₙ = a₁ + (n – 1)d.
a₁₇ = 116 +(17-1)(-3)
a₁₇ = 116+(16)(-3)
a₁₇ = 116 - 48
a₁₇ = 68
So, the 17th term in arithmetic sequence is 68
You might be able to retake the quarter test but i dont think you wont pass 7th grade.
For this case we must find the product of the following expressions:
\frac {3x} {x + 1} * \frac {x} {x-7}
So:
\frac {3x ^ 2} {x ^ 2-7x + x-7} =\\\frac {3x ^ 2} {x ^ 2-6x-7}
So, we have to:
\frac {3x} {x + 1} * \frac {x} {x-7} = \frac {3x ^ 2} {x ^ 2-6x-7}
Answer:
\frac {3x ^ 2} {x ^ 2-6x-7}
Step-by-step explanation:
