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3 years ago

13

Write an equation for the nth term of the geometric sequence below: (Equation NEEDS to be written as gn=7(5)^(n-1) or as gn=2/3(

1/4)^(n-1) NOTICE NO SPACES & using ^ to represent an exponent)

1 answer:

andriy [413]3 years ago

8 0

Answer:

You didn't give the geometric sequence in question.

But to find the nth term of a geometric sequence, use the formula

Sn = ar^(n - 1)

Where a is the first term of the sequence

r is the common ratio of the sequence.

Example: To find the nth term of the geometric sequence

1/2, 1/4, 1/8, 1/16, ...

Here, a = 1/2

r = 1/4 ÷ 1/2 = 1/8 ÷ 1/4 = 1/16 ÷ 1/8 = 1/2.

Sn = (1/2)(1/2)^( n - 1)

Apply this method to the sequence you were given, it will be helpful.

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7. Jena has a triangular garden with a perimeter that can be represented by 25r ^ 2 - 9r + 14 . The sum of two sides of the gard

galben [10]

Answer: $15r^2-21r+5$

Step-by-step explanation:

Given: Perimeter of triangular garden = $25r^2 - 9r + 14$

The sum of two sides of the garden = $10r^2 + 12r + 9$

Perimeter is the sum of all sides of a triangle.

Third side = Perimeter of triangular garden - sum of two sides of the garden

$=25r ^ 2 - 9r + 14-(10r ^ 2 + 12r + 9)\\\\=25r ^ 2 - 9r + 14-10r ^ 2 - 12r - 9\\\\=15r^2-21r+5$

The expression represent the third side = $15r^2-21r+5$ .

5 0

3 years ago

SOLVE THE QUESTION BELOW ASAP

qwelly [4]

Answer:

Part A) The graph in the attached figure (see the explanation)

Part B) 16 feet

Part C) see the explanation

Step-by-step explanation:

Part A) Graph the function

Let

h(t) ----> the height in feet of the ball above the ground

t -----> the time in seconds

we have

$h(t)=-16t^{2}+98$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex is a maximum

To graph the parabola, find the vertex, the intercepts, and the axis of symmetry

<em>Find the vertex</em>

The function is written in vertex form

so

The vertex is the point (0,98)

Find the y-intercept

The y-intercept is the value of the function when the value of t is equal to zero

For t=0

$h(t)=-16(0)^{2}+98$

$h(0)=98$

The y-intercept is the point (0,98)

Find the t-intercepts

The t-intercepts are the values of t when the value of the function is equal to zero

For h(t)=0

$-16t^{2}+98=0$

$t^{2}=\frac{98}{16}$

square root both sides

$t=\pm\frac{\sqrt{98}}{4}$

$t=\pm7\frac{\sqrt{2}}{4}$

therefore

The t-intercepts are

$(-7\frac{\sqrt{2}}{4},0), (7\frac{\sqrt{2}}{4},0)$

$(-2.475,0), (2.475,0)$

Find the axis of symmetry

The equation of the axis of symmetry in a vertical parabola is equal to the x-coordinate of the vertex

so

$x=0$ ----> the y-axis

To graph the parabola, plot the given points and connect them

we have

The vertex is the point $(0,98)$

The y-intercept is the point $(0,98)$

The t-intercepts are $(-2.475,0), (2.475,0)$

The axis of symmetry is the y-axis

The graph in the attached figure

Part B) How far is the artifact fallen from the time t=0 to time t=1

we know that

For t=0

$h(t)=-16(0)^{2}+98$

$h(0)=98\ ft$

For t=1

$h(t)=-16(1)^{2}+98$

$h(1)=82\ ft$

Find the difference

$98\ ft-82\ ft=16\ ft$

Part C) Does the artifact fall the same distance from time t=1 to time t=2 as it does from the time t=0 to time t=1?

we know that

For t=1

$h(t)=-16(1)^{2}+98$

$h(1)=82\ ft$

For t=2

$h(t)=-16(2)^{2}+98$

$h(2)=34\ ft$

Find the difference

$82\ ft-34\ ft=48\ ft$

so

The artifact fall 48 feet from time t=1 to time t=2 and fall 16 feet from time t=0 to time t=1

therefore

The distance traveled from t=1 to t=2 is greater than the distance traveled from t=0 to t=1

8 0

3 years ago

(-5-2)x(-3) what is the answer

fenix001 [56]

Answer:

21

Step-by-step explanation:

-5 - 2= -7

-7*-3=

<h2>21</h2>

7 0

4 years ago

Read 2 more answers

HELP PLEASE!! thanks

Vesna [10]

Answer:

352

but for the 3 it would be 264

5 0

3 years ago

How to calculate 99 : 5

Goryan [66]

Answer:

19.8

Step-by-step explanation:

99/5= 19.8

5 0

3 years ago