20 people show up to the party, and Robert decides to give them each the same

Find the common difference of the following sequence.

Darya [45]

You have the correct answer. It is choice B) -1/4

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Explanation:

This is because we're adding -1/4 to each term to get the next one. In other words, we're subtracting 1/4 from each term to get the next one.

term2 = term1+(d) = 1/2 + (-1/4) = 1/2 - 1/4 = 2/4 - 1/4 = 1/4
term3 = term2+(d) = 1/4 + (-1/4) = 1/4 - 1/4 = 0
term4 = term3+(d) = 0 + (-1/4) = 0 - 1/4 = -1/4
term5 = term4+(d) = -1/4 + (-1/4) = -2/4 = -1/2

and so on.

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To find the common difference, all we have to do is subtract any term from its previous one.

For example:

d = (term2) - (term1)

d = (1/4) - (1/2)

d = (1/4) - (2/4)

d = (1-2)/4

d = -1/4

The order of subtraction matters, so we cannot say d = term1-term2.

3 0

3 years ago

Two-fifths of the senior class earned a grade of B+ or higher on an advanced mathematics exam.

matrenka [14]

2/5 = 42
1/5 = 21
5/5 = 105

105 students in the class

5 0

3 years ago

Read 2 more answers

Find a function where f(0)=2 and f(1)=2

xenn [34]

Answer:

Do you want to be extremely boring?

Since the value is 2 at both 0 and 1, why not make it so the value is 2 everywhere else?

$f(x) = 2$ is a valid solution.

Want something more fun? Why not a parabola? $f(x)= ax^2+bx+c$ .

At this point you have three parameters to play with, and from the fact that $f(0)=2$ we can already fix one of them, in particular $c=2$ . At this point I would recommend picking an easy value for one of the two, let's say $a= 1$ (or even $a=-1$ , it will just flip everything upside down) and find out b accordingly: $f(1)=2 \rightarrow 1^2+b+2=2 \rightarrow b=-1$

Our function becomes

$f(x) = x^2-x+2$

Notice that it works even by switching sign in the first two terms: $f(x) = -x^2+x+2$

Want something even more creative? Try playing with a cosine tweaking it's amplitude and frequency so that it's period goes to 1 and it's amplitude gets to 2: $f(x) = A cos (kx)$

Since cosine is bound between -1 and 1, in order to reach the maximum at 2 we need $A= 2$ , and at that point the first condition is guaranteed; using the second to find k we get $2= 2 cos (k1) = cos k = 1 \rightarrow k = 2\pi$