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3 years ago

Given the following recurrence relation: An=(−2)An−1, with A0=(−1)

1 answer:

4 0

Technically you could answer this one by one, by substituting n with 1, 2, 3, 4, then 5. Like here:
A1=(-2)A1-1
=(-2)A0, then substituting A0 with -1...
=(-2)(-1)
Which would mean A1=2. Then..
A2=(-2)A2-1
A2=(-2)A1
A2=(-2)(2)
Which would mean A2=-4.
By repeating this process you'd eventually get A4..
A5=(-2)A4
if you did the steps right you'd get that A4 is -16..
A5=(-2)(-16).
Then simplify to get your answer.

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Help please!!! photo above

Viktor [21]

Answer:

5<*line under* x >*line under*9

I think

4 0

3 years ago

A triangle has a base of 54 cm. The height is unknown. Complete the equation that can be used to find the area, A, of the triang

Kay [80]

Answer:

A=1/2× base ×hieght

A=1/2×b×h

6 0

3 years ago

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maria [59]

I think the answer is 9
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3 0

3 years ago

Solve for y.<br> r/3-2/y=s/5

nordsb [41]

Answer:

y = 2 / (r/3 - s/5)

Step-by-step explanation:

r/3 - 2/y = s/5

add 2/y to both sides

r/3 = s/5 + 2/y

Subtract s/5 from both sides

r/3 - s/5 = 2/y

multiply both sides by y

y(r/3 - s/5) = 2

Divide both sides by r/3 - s/5

y = 2 / (r/3 - s/5)

8 0

3 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

4 years ago