Answer:
x > 1
Step-by-step explanation:
Given
8x - 2 > 6 ( isolate the term in x by adding 2 to both sides )
8x > 8 ( divide both sides by 8 )
x > 1
Answer:
x axis
Step-by-step explanation:
We know the x value stays the same. If you were to graph the line y=3 and y=-3, you would see that it is a reflection over the x axis.
Answer:
14x + 8
Explanation:
⇒ 4(5x+5) - 3(2x + 4)
distribute inside parenthesis
⇒ 4(5x) + 4(5) - 3(2x) - 3(4)
multiply the variables
⇒ 20x + 20 - 6x - 12
collect like terms
⇒ 20x - 6x + 20 - 12
subtract like term
⇒ 14x + 8
Hey!
To solve this equation, we'd first have to apply the distributing rule to this equation.
<em>Original Equation :</em>
<em>New Equation {Changed by Applying the Distribution Rule} :</em>
· 3a · 2 +
And now we simplify.
<em>Old Equation :</em>
· 3a · 2 +
<em>New Equation {Old Equation Simplified} :</em>
- 12a + 4
<em>So, the equation
simplified is</em>
- 12a + 4.Hope this helps!
- Lindsey Frazier ♥
Since we are already given the amount of jumps from the first trial, and how much it should be increased by on each succeeding trial, we can already solve for the amount of jumps from the first through tenth trials. Starting from 5 and adding 3 each time, we get: 5 8 (11) 14 17 20 23 26 29 32, with 11 being the third trial.
Having been provided 2 different sigma notations, which I assume are choices to the question, we can substitute the initial value to see if it does match the result of the 3rd trial which we obtained by manual adding.
Let us try it below:
Sigma notation 1:
10
<span> Σ (2i + 3)
</span>i = 3
@ i = 3
2(3) + 3
12
The first sigma notation does not have the same result, so we move on to the next.
10
<span> Σ (3i + 2)
</span><span>i = 3
</span>
When i = 3; <span>3(3) + 2 = 11. (OK)
</span>
Since the 3rd trial is a match, we test it with the other values for the 4th through 10th trials.
When i = 4; <span>3(4) + 2 = 14. (OK)
</span>When i = 5; <span>3(5) + 2 = 17. (OK)
</span>When i = 6; <span>3(6) + 2 = 20. (OK)
</span>When i = 7; 3(7) + 2 = 23. (OK)
When i = 8; <span>3(8) + 2 = 26. (OK)
</span>When i = 9; <span>3(9) + 2 = 29. (OK)
</span>When i = 10; <span>3(10) + 2 = 32. (OK)
Adding the results from her 3rd through 10th trials: </span><span>11 + 14 + 17 + 20 + 23 + 26 + 29 + 32 = 172.
</span>
Therefore, the total jumps she had made from her third to tenth trips is 172.
