Answer:
= 5n
Step-by-step explanation:
There is a common difference d between consecutive terms
d = 10 - 5 = 15 - 10 = 20 - 15 = 25 - 20 = 30 - 25 = 5
This indicates the sequence is arithmetic with explicit formula
= a₁ + (n - 1)d
where a₁ is the first term and d the common difference
Here a₁ = 5 and d = 5 , then
= 5 + 5(n - 1) = 5 + 5n - 5 = 5n
19. What you know is that HK+KJ = HJ. If HJ = 25, the sum of the two equations will equal this length.
x-5+5x-12=25 First, combine your like terms. You will end up with 6x-17=25. Add the opposite of -17 to both sides. 6x = 42 Divide both sides by 6. x = 7. Substitute x=7 for your original expression of x-5, 7-5=2
20. (5x-6)/2 = x+6 Multiply each side by 2. 5x-6 = 2x +12 Add 6 to each side 5x = 2x + 18 then subtract 2x from both sides as well. 3x = 18 Finally divide each side by 3. x=6 To find the length of the remaining segment, substitute this value into (5x-6)/2. This results in each side equaling a distance of 12.
21. On the number line, the distance of FG is 16 units. If the distance of FP is 1/4 of FG, you would simply divide 16 by 4. The distance of FP is 4 and P lies at 8 on your number line.
23. The distance of SP is x+4 and ST=4x. Since P is the midpoint, you only have one half of the line as x+4, if you were to double it, you would find that 2x+8 = 4x. Balance and solve for x, subtract 2x from both sides. 8=2x Divide each side by 4, 8/4 = 4x/4 resulting in x=2. If ST equals 4x, substitute and solve, 4(2) = 8
60/150 = 40%
Hope this helps :D
It's clear that for x not equal to 4 this function is continuous. So the only question is what happens at 4.
<span>A function, f, is continuous at x = 4 if
</span><span>
</span><span>In notation we write respectively
</span>
Now the second of these is easy, because for x > 4, f(x) = cx + 20. Hence limit as x --> 4+ (i.e., from above, from the right) of f(x) is just <span>4c + 20.
</span>
On the other hand, for x < 4, f(x) = x^2 - c^2. Hence
Thus these two limits, the one from above and below are equal if and only if
4c + 20 = 16 - c²<span>
Or in other words, the limit as x --> 4 of f(x) exists if and only if
4c + 20 = 16 - c</span>²
That is to say, if c = -2, f(x) is continuous at x = 4.
Because f is continuous for all over values of x, it now follows that f is continuous for all real nubmers 
Answer:
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