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Tema [17]
3 years ago
13

The seats at a local baseball stadium are arranged so that each row has 5 more seat than the row in front of it. If there are fo

ur seats in the first row, how many total seats are in the first 24 rows
Mathematics
1 answer:
ohaa [14]3 years ago
4 0
You would do an arithmetic sequence
use A1 +(n-1)d to calculate number of seats in the last row

An = 4+(24-1)5
An = 4 + (23)(5)
An = 4 + 115
An = 119 seats in the last row

now use a sum sequence to find total number of seats

Sn = n(a1 + an)/2 
Sn = 24(4+119)/2 
Sn = 24(123) /2 
Sn = 1476 total seats

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Jase's farm has 9 chickens, each of which laid 3 eggs. Jases farm has 4 horses. Edna's farm has chickens which laid a total of 2
s344n2d4d5 [400]

Answer:

You can use the basic multipication as an infrence



5 0
3 years ago
Choose the equation you would use to find the altitude of the airplane. tan70 = 0 tan70 = sin70 =
KonstantinChe [14]
The situation is represented by a drawing with a right triangle where an airplane is at the top corner, the elevation angle (opposite to the airplane) is 70°, the height (vertical leg) is x, and the adjacent leg (horizontal leg) to the 70° angle is 800.

Then, to find the height x, which is the opposite leg to the angle,  you can use the tangent ratio, which is opposite leg divided by adjacent leg:

tan(x) = opposite leg / adjacent leg => tan(70°) = x / 800

Answer: tan(70°) = x / 800
7 0
3 years ago
When the effective interest rate is 9% per annum, what is the present value of a series of 50 annual payments that start at $100
ser-zykov [4K]

Answer:

$1,109.62

Step-by-step explanation:

Let's first compute the <em>future value FV.</em>  

In order to see the rule of formation, let's see the value (in $) for the first few years

<u>End of year 0</u>

1,000

<u>End of year 1(capital + interest + new deposit)</u>

1,000*(1.09)+10  

<u>End of year 2 (capital + interest + new deposit)</u>

(1,000*(1.09)+10)*1.09 +10 =

\bf 1,000*(1.09)^2+10(1+1.09)

<u>End of year 3 (capital + interest + new deposit)</u>

\bf (1,000*(1.09)^2+10(1+1.09))(1.09)+10=\\1,000*(1.09)^3+10(1+1.09+1.09^2)

and we can see that at the end of year 50, the future value is

\bf FV=1,000*(1.09)^{50}+10(1+1.09+(1.09)^2+...+(1.09)^{49}

The sum  

\bf 1+1.09+(1.09)^2+...+(1.09)^{49}

is the <em>sum of a geometric sequence </em>with common ratio 1.09 and is equal to

\bf \frac{(1.09)^{50}-1}{1.09-1}=815.08356

and the future value is then

\bf FV=1,000*(1.09)^{50}+10*815.08356=82,508.35564

The <em>present value PV</em> is

\bf PV=\frac{FV}{(1.09)^{50}}=\frac{82508.35564}{74.35572}=1,109.616829\approx \$1,109.62

rounded to the nearest hundredth.

5 0
3 years ago
Write an equation for 14 groups of 16
ZanzabumX [31]
14 times 16=84+140=224
4 0
3 years ago
Question #2 how do i solve this.
bulgar [2K]
It could be solved by analyzing the questions and finding important words that stand out to you. 20=16+4 , 50=10+40. this is the way you find it
6 0
4 years ago
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