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UkoKoshka [18]
3 years ago
10

Which of the following sequences is not arithmetic?

Mathematics
1 answer:
Reil [10]3 years ago
7 0

Answer:

B.

Step-by-step explanation:

Arithmetic sequences have a definite pattern.

A is counting up by 1, C is counting up by 2, and D is subtracting by 2 to get the next number in their sequences.

B's pattern is not so definite as it jumps from 14 to 9 and then to 16.

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Choose all that define a rectangle in the coordinate plane by the four given points.
inysia [295]

Answer:

The answer is A(-3,-4) B(-1,2) C(2,1) D(0,-5)

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Describe the graph of a function g by observing the graph of the base function f ? ​
never [62]

Answer:

Choice 3 is your answer

Step-by-step explanation:

The format of the function when you move it side to side or up and down is

f(x) = (x - h) + k,

where h is the side to side movement and k is up or down.  The k is easy, since it will be positive if we move the function up and negative if we move the function down from its original position.

The h is a little more difficult, but just remember the standard form of the side to side movement is always (x - h).  If our function has moved 3 units to the left, we fit that movement into our standard form as (x - (-3)), which of course is the same as (x + 3).  Our function has moved up 5 units, so the final translation is

g(x) = f(x + 3) + 5, choice 3 from the top.

5 0
3 years ago
Let y(????)y(t) be a solution of y˙=17y(1−y7)y˙=17y(1−y7) such that y(0)=14y(0)=14. Determine lim????→[infinity]y(????)limt→[inf
shtirl [24]

Answer:

Step-by-step explanation:

Given that,

y' = 17y ( 1-y^7)

Let y=1

Then, y' = 0 for all t

Then show that it is the only stable equilibrium point so that as y→1, t→∞ with any initial value.

So, the graph solution will be

y(0) = 1 and this will be an horizontal line

If, y(0) > 1 then, y' < 0 by inspecting the first equation, so the graph is has decreasing solution.

Likewise, if y(0) < 1 then, y' > 0 and the graph is increasing.

So no matter the initial condition, graph of the solution will be asymptotic to the horizontal line above.

This make the limit be 1.

This shows that x = 1 is a stable equilibrium.

6 0
3 years ago
An experiment involves 30 participants. From these, a group of 4 participants is to be tested under a special condition. How man
zheka24 [161]

Answer:

7

Step-by-step explanation:

7*4=28 which is the closest u can get to 30 with out going past it

6 0
3 years ago
Suppose that 15 inches of wire costs 90 cents. At the same rate, how many inches of wire can be bought for 48 cents?
tamaranim1 [39]
This can be solved by making an equivalent ratio.
The original ratio is what we know, 15 inches of wire for 90 cents.
In a ratio of inches of wire:cents, this would be 15:90.

Now for the equivalent ratio.
We don't know the number in the inches place but we do know it for the cents place.
Let's use x to represent inches of wire.
x:48 is our new ratio, and we need to find x.

Since x:48 and 15:90 are equivalent, that means the same thing that was done to 90 to get 48 has to be done to 15 to get the value of x, since the same thing must be applied to both sides.
We can find what 90 was divided by (which is what we'll have to divide 15 by) by dividing 90 by 48.
90 / 48 = 1.875

This means 48 • 1.875 = 90 and x • 1.875 = 15.
Since we don't know x though, we can isolate it by dividing both sides by 1.875.
x • 1.875 = 15
x • 1.875 / 1.875 = x
15 / 1.875 = 8
So x is 8.

Answer:
While you can be 15 inches of wire for 90 cents, you can buy 8 inches of wire for 48 cents at the same rate.
3 0
3 years ago
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