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3 years ago

These are the first six terms of a sequence with a = 2:

Mathematics

2 answers:

soldi70 [24.7K]3 years ago

6 0

Answer: The formular for this sequence is AR^n-1 (that is, A multiplied by R{raised to the power of n minus 1} )

Step-by-step explanation:This is a geometric progression in which every term is calculated by multiplying each previous term by a common ratio.

The common ratio here is 7, which is derived as

14/2, or 98/14, or 686/98, or 4802/686...

In simply put, R is derived as Tn/Tn-1, where Tn is the nth term and Tn-1 is the previous term.

Therefore the formular for this progression is given as

AR^n-1

Where A = 2, R = 7 and n = the nth term.

SpyIntel [72]3 years ago

3 0

Answer:

Step-by-step explanation:

98÷14

686÷98

4 802÷686

33 614÷4 802

Then the common ratio q for this sequence is 7

recursive formula : an+1 = q×an = ?

an= a1 × qⁿ⁻¹

=2×7ⁿ⁻¹

an+1 = q×an

= 7×(2×7ⁿ⁻¹)

= 2×7ⁿ

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DENIUS [597]

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3 years ago

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DaniilM [7]

1. Remember that the perimeter is the sum of the lengths of the sides of a figure.To solve this, we are going to use the distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
where
$(x_{1},y_{1})$ are the coordinates of the first point
$(x_{2},y_{2})$ are the coordinates of the second point
Length of WZ:
We know form our graph that the coordinates of our first point, W, are (1,0) and the coordinates of the second point, Z, are (4,2). Using the distance formula:
$d_{WZ}= \sqrt{(4-1)^2+(2-0)^2}$
$d_{WZ}= \sqrt{(3)^2+(2)^2}$
$d_{WZ}= \sqrt{9+4}$
$d_{WZ}= \sqrt{13}$

We know that all the sides of a rhombus have the same length, so
$d_{YZ}= \sqrt{13}$
$d_{XY}= \sqrt{13}$
$d_{XW}= \sqrt{13}$

Now, we just need to add the four sides to get the perimeter of our rhombus:
$perimeter= \sqrt{13} + \sqrt{13} + \sqrt{13} + \sqrt{13}$
$perimeter=4 \sqrt{13}$
We can conclude that the perimeter of our rhombus is $4 \sqrt{13}$ square units.

2. To solve this, we are going to use the arc length formula: $s=r \alpha$
where
$s$ is the length of the arc.
$r$ is the radius of the circle.
$\alpha$ is the central angle in radians

We know form our problem that the length of arc PQ is $\frac{8}{3} \pi$ inches, so $s=\frac{8}{3} \pi$ , and we can infer from our picture that $r=15$ . Lest replace the values in our formula to find the central angle POQ:
$s=r \alpha$
$\frac{8}{3} \pi=15 \alpha$
$\alpha = \frac{\frac{8}{3} \pi}{15}$
$\alpha = \frac{8}{45} \pi$

Since $\alpha =POQ$ , We can conclude that the measure of the central angle POQ is $\frac{8}{45} \pi$

3. A cross section is the shape you get when you make a cut thought a 3 dimensional figure. A rectangular cross section is a cross section in the shape of a rectangle. To get a rectangular cross section of a particular 3 dimensional figure, you need to cut in an specific way. For example, a rectangular pyramid cut by a plane parallel to its base, will always give us a rectangular cross section.
We can conclude that the draw of our cross section is: