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Ivahew [28]
3 years ago
11

Determine whether the sequence is arithmetic or geometric. Sequence 1: –10, 20, – 40, 80, ... Sequence 2: 15, – 5, – 25, – 45, .

.. Which of the following statements are true regarding Sequence 1 and Sequence 2.
Mathematics
1 answer:
harina [27]3 years ago
7 0

Answer:

geometric

arithmetic

Step-by-step explanation:

there are no statements to check. you did not copy them in.

but I can solve the first part of the question.

sequence 1 :

-10, 20, -40, 80, -160, 320, -640, ...

I extended the sequence by a few more terms.

but it is clear that each new term is the previous term multiplied by -2.

since the sequence is built by multiplication, it is geometric.

sequence 2:

15, -5, -25, -45, -65, -85, ...

so, it is clear that each new term is the previous term subtracted by -20.

since the sequence is built by adding or subtracting, it is arithmetic.

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The value 170 000 is 100x larger than the value 170.
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Which of the following is a solution of z^5 = 1 + √3 i?
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Answer:

Option 2 is right

Step-by-step explanation:

Given that

z^5=1+\sqrt{3} i

We can write this in polar form with modulus and radius

|z^5|= \sqrt{1+3} =2\\tan of Angle t =\sqrt{3} \\

Hence angle = 60 degrees and

|z^5|= 2(cos60+isin60)

Since we have got 5 roots for z, we can write 60, 420, 780, etc. with periods of 360

Using Demoivre theorem we get 5th root would be

5th root of 2 multiplied by 1/5 th of 60, 420, 780,....

z= \sqrt[5]{2} (cos12+isin12)\\z=\sqrt[5]{2} (cos84+isin84)\\\\z=\sqrt[5]{2} (cos156+isin156)\\\\z=\sqrt[5]{2} (cos228+isin228)\\\\z=\sqrt[5]{2} (cos300+isin300)\\

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3 years ago
30 POINTS PLEASE HELP!!! 4. The following equations represent the same quadratic function written in standard, vertex, and inter
sergij07 [2.7K]

Answer:

A

Step-by-step explanation:

So we have the quadratic equation and it's written in three equivalent forms:

f(x)=0.5x^2+x-1.5\\f(x)=0.5(x+1)^2-2\\f(x)=(0.5x+1.5)(x-1)

Let's determine the characteristics of the quadratic equation with the given equations.

From the first equation, since the leading coefficient (0.5) is positive, we can be certain that the graph opens upwards.

Also, the constant term is -1.5, so the y-intercept is (0,-1.5).

The second equation is the vertex form. Vertex form has the format:

f(x)=a(x-h)^2-k

Where (h,k) is the vertex. From the second equation we know that h is -1 (because (x+1) is the same as (x-(-1))) and k is -2. Therefore, the vertex is (-1,-2).

And since the graph points upwards, this means that (-1,-2) is the minimum point of the function. In other words, the range of the function is greater than or equal to -2. In interval notation, this is:

[-2,\infty)

This also means that the end behavior of the graph as a x approaches negative and positive infinity is positive infinity because the graph will always go straight up.

Also, the third form is the factored form. With that, we can solve for the zeros of the quadratic. The zeros are:

0.5x+1.5=0\text{ and } x-1=0\\0.5x=-1.5 \text{ and }x=1\\x=-3\text{ and }x=1

Therefore, the graph crosses the x-axis at x=-3 and x=1.

So, from the three equations, we gathered the following information:

1) The graph curves upwards.

2) The roots of zeros of the function is (-3,0) and (1,0).

3) The y-intercept is (0,-1.5).

4) The vertex is (-1,-2). This is also the minimum point.

5) Therefore, the range of the graph is all values greater than or equal to -2.

6) The end behavior of the graph on both directions go towards positive infinity.

Therefore, our correct answer is A.

B is not correct because the line of symmetry (or the x-coordinate of the vertex) here is -1 and not 1/2.

C is not correct because the graph goes towards <em>positive </em>infinity since it shoots straight up.

And D is not correct because the y-intercept is (0,-1.5).

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