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ss7ja [257]
3 years ago
6

1. In an auditorium, there are 21 seats in the first row and 26 seats in the second row. The number of seats in a row continues

to increase by 5 with each additional row.
(a) Write an iterative (explicit) rule to model the sequence formed by the number of seats in each row. Show your work.

(b) Use the rule to determine how many seats are in row 15. Show your work.


2. Rhonda started a business. Her business made $40,000 in profits the first year. Her annual profits have increased by an average of 6% each year since then.

(a) Write an iterative rule to model the sequence formed by the profits of Rhonda’s business each year.

(b) Use the rule to determine what the annual profits of Rhonda’s business can be predicted to be 20 years from the start of her business. Round your answer to the nearest dollar. Do not round until the end. Show your work.

3. The sequence 3, 12, 48, 192, … shows the number of pushups Kendall did each week, starting with her first week of exercising.

(a) What is the recursive rule for the sequence?

(b) What is the iterative rule for the sequence?
Mathematics
1 answer:
kondor19780726 [428]3 years ago
6 0
1. Let s_n be the number of seats in the n-th row. The number seats in the n-th row relative to the number of seats in the (n-1)-th row is given by the recursive rule

s_n=s_{n-1}+5


Since s_1=21, we have

s_2=s_1+5
s_3=s_2+5=s_1+2\cdot5
s_4=s_3+5=s_1+3\cdot5
\cdots
s_n=s_{n-1}+5=\cdots=s_1+(n-1)\cdot5

So the explicit rule for the sequence s_n is

s_n=21+5(n-1)\implies s_n=5n+16

In the 15th row, the number of seats is


s_{15}=5(15)+16=91

2. Let p_n be the amount of profit in the n-th year. If the profits increase by 6% each year, we would have

p_2=p_1+0.06p_1=1.06p_1
p_3=1.06p_2=1.06^2p_1
p_4=1.06p_3=1.06^3p_1
\cdots
p_n=1.06p_{n-1}=\cdots=1.06^{n-1}p_1

with p_1=40,000.

The second part of the question is somewhat vague - are we supposed to find the profits in the 20th year alone? the total profits in the first 20 years? I'll assume the first case, in which we would have a profit of


p_{20}=1.06^{19}\cdot40,000\approx121,024

3. Now let p_n denote the number of pushups done in the n-th week. Since 3\cdot4=12, 12\cdot4=48, and 48\cdot4=192, it looks like we can expect the number of pushups to quadruple per week. So,

p_n=4p_{n-1}

starting with p_1=3.

We can apply the same reason as in (2) to find the explicit rule for the sequence, which you'd find to be

p_n=4^{n-1}p_1\implies p_n=4^{n-1}\cdot3
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All you have to do is plug in.

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6 0
3 years ago
Roger the park ranger was standing in a look out tower that was 200 ft tall. Looks down, sees a fire. Angle of depression is 7 d
Ainat [17]

Answer:

approx. 24.6 ft

Step-by-step explanation:

x=the distance of the fire from the base of the tower

200ft=the height of the tower

7 degrees=the angle of depression

Using these values you can draw a diagram that highlights that you need to use trig. to work out the answer.

Using Tan:

Tan 7=x/200 ft

multiply both sides by 200 ft

200*tan 7=24.55691218

approx. 24.6 ft

8 0
3 years ago
Ivan is painting 3 walls in his living room. Each wall measures 9 3/4 feet tall by 14 1/4 feet wide. Ivan needs to estimate the
Alenkinab [10]

Answer:

The estimated area is 416\frac{13}{16}\ ft^{2}

Step-by-step explanation:

we know that

To estimate the area that Ivan will paint , calculate the area of one wall and then multiply by 3

step 1

Convert the dimensions to an improper fractions

9\frac{3}{4}\ ft=\frac{9*4+3}{4}=\frac{39}{4}\ ft

14\frac{1}{4}\ ft=\frac{14*4+1}{4}=\frac{57}{4}\ ft

step 2

Find the area of one wall

(\frac{39}{4})*(\frac{57}{4})=\frac{2,223}{16}\ ft^{2}

step 3

Multiply the area of one wall by 3

(3)*\frac{2,223}{16}=\frac{6,669}{16}\ ft^{2}

step 4

Convert to mixed number

\frac{6,669}{16}=\frac{6,656}{16}+\frac{13}{16}=416\frac{13}{16}\ ft^{2}

4 0
3 years ago
The sea ice area around the North Pole fluctuates between about 6 million square kilometers in September to 14 million square ki
Aleksandr [31]

Answer:

there are approximately 5.035 months when there is less than 9 million square meters of sea ice around the North Pole in a year.

Step-by-step explanation:

Given the data in the question;

Let S(t) represent the amount sea ice around the North Pole in millions of square meters at a given time t,

t is the number of months since January.

Now, we use a cosine curve to model this scenario

Vertical shift will be;

D = ( 6 + 14 ) / 2 = 20 / 2

D = 10

Next is the Amplitude;

|A| = ( 6 - 14 ) / 2

|A| = 4

Now, the horizontal stretch factor will be;

B = 2π / 12

B = π/6

Hence;

S(t) = 4cos( π/6 × t ) + 10 ----------- let this be equation 1

Now we find when there will be less than 9 million square meters of sea ice;

S(t) = 9

so we have

9 = 4cos( π/6 × (t-2) ) + 10

9 - 10 = 4cos( π/6 × (t-2) )

-1 = 4cos( π/6 × (t-2) )

-1/4 = cos( π/6 × (t-2) )

so we have;

cos⁻¹( -1/4 ) = π/6 × (t₁-2)  -------- let this be equation 2

2π - cos⁻¹( -1/4 ) = π/6 × (t₂-2)  -------- let this be equation 3

so we solve equation 2 and 3

we have'

t₁ - t₂ = 6/π × ( 2π - cos⁻¹( -1/4 ) - cos⁻¹( -1/4 ) )

t₁ - t₂ = 6/π × ( 2π - 2cos⁻¹( -1/4 )  

t₁ - t₂ = 6/π × ( π - cos⁻¹( -1/4 )  

t₁ - t₂ = 6/π × ( π - 104.4775 )

t₁ - t₂ = 6/π × ( π - 104.4775 )  

t₁ - t₂ = 5.035

therefore, there are approximately 5.035 months when there is less than 9 million square meters of sea ice around the North Pole in a year.

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3 years ago
List 3 rational numbers between 3 and 3.9
antiseptic1488 [7]

Answer:

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Step-by-step explanation:

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