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2 years ago

A geometric seqerence 1.5, -3, 6, -12 .... How many regative terms in the sequencs greater than

1 answer:

4 0

The total negative terms are 4 which are greater than -6000 if the geometric sequence 1.5, -3, 6, -12 ... with a common ratio -3.

<h3>What is a sequence?</h3>

It is defined as the systematic way of representing the data that follows a certain rule of arithmetic.

We have:

A geometric sequence 1.5, -3, 6, -12 ....

The common ratio r = -3/1.5 = -2

1.5, -3.6, 6, -12, 36, -108, 324, -972, 2916, -8748

The total negative terms = 4

Thus, the total negative terms are 4 which are greater than -6000 if the geometric sequence 1.5, -3, 6, -12 ... with a common ratio -3.

Learn more about the sequence here:

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If the 10 is being multiplied it’s 90

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Therefore, there is information missing in order to solve the question

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3 years ago

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3 years ago

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sattari [20]

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2 years ago

Is the given point interior , exterior, or on the circle k (x+2)2 + (y-3)2 =18 P (8,4)

Mnenie [13.5K]

One way would be to find the distance from the point to the center of the circle and compare it to the radius

for
$(x-h)^2+(y-k)^2=r^2$
the center is (h,k) and the radius is r

and the distance formula is
distance between $(x_1,y_1)$ and $(x_2,y_2)$ is
$D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

r=radius
D=distance form (8,4) to center

if r>D, then (8,4) is inside the circle
if r=D, then (8,4) is on the circle
if r<D, then (8,4) is outside the circle

so
$(x+2)^2+(y-3)^2=18$
$(x-(-2))^2+(y-3)^2=(\sqrt{18})^2$
$(x-(-2))^2+(y-3)^2=(3\sqrt{2})^2$

the radius is $3\sqrt{2}$
center is (-2,3)

find distance between (8,4) and (-2,3)

$D=\sqrt{(8-(-2))^2+(4-3)^2}$
$D=\sqrt{(8+2)^2+(1)^2}$
$D=\sqrt{10^2+1}$
$D=\sqrt{100+1}$
$D=\sqrt{101}$

$r=3\sqrt{2}$ ≈4.2
$D=\sqrt{101}$ ≈10.04

do r<D

(8,4) is outside the circle

6 0

3 years ago