One solution to the problem below is 4. What is the other solution? y² - 16 = 0
1 answer:
You might be interested in
The point of intersection is the point where lines intersect.
<em>There will be 595 intersections for 35 lines, where no 3 lines are concurrent.</em>
<em />
Given
<em /><em> --- the number of lines</em>
<em /><em> --- no three lines are concurrent</em>
<em />
When no three line are concurrent, it means that no three lines meet at the same point.
<u>So, the sequence of intersection is:</u>
- <em>0 intersection for 1 line</em>
- <em>1 intersection for 2 lines</em>
- <em>3 intersections for 3 lines</em>
- <em>6 intersections for 4 lines</em>
<em />
Following the above sequence, the number of intersections for n lines is:
In this case,
So, we have:
<em>Hence, there will be 595 intersections for 35 lines, where no 3 lines are concurrent.</em>
<em />
Read more about lines of intersections at:
The first year, the amount is 40,000
the second year is 40000 + 4.2% of 40000, or 0.042 * 4000, so 40000+(0.042*4000)
common factoring that we get 40000(1 + 0.042), or just 40000(1.042)
in short, the starting amount is 40000, and to get the next term's value you'd use the "common ratio" of 1.042, namely the multiplier of 1.042.
for the third year it'll be 40000(1.042) + (0.042 *40000(1.042) ), again, common factoring that
40000(1.042)(1 + 0.042) or 40000(1.042)(1.042) or 40000(1.042)²
therefore,
the second year is 40000 + 4.2% of 40000, or 0.042 * 4000, so 40000+(0.042*4000)
common factoring that we get 40000(1 + 0.042), or just 40000(1.042)
in short, the starting amount is 40000, and to get the next term's value you'd use the "common ratio" of 1.042, namely the multiplier of 1.042.
for the third year it'll be 40000(1.042) + (0.042 *40000(1.042) ), again, common factoring that
40000(1.042)(1 + 0.042) or 40000(1.042)(1.042) or 40000(1.042)²
therefore,
Answer:
x = -10
Step-by-step explanation:
We just have to translate the problem correctly
>8 times a number is the same
this tells me that
>as 30 less than 5 times the number.
Now set them equal to eachother
Now isolate the variable x