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Natasha2012 [34]
3 years ago
6

A small auditorium has 10 rows of seats. There are 12 seats in the row and 16 seats in the second row the number of seats in a r

ow continues to increase by 4 with each additional row. What is the total number of seats in the auditorium?
Mathematics
1 answer:
guapka [62]3 years ago
4 0

This problem can be solved through simple arithmetic progression

Let

a1 = the first term of the sequence

a(n) = the nth term of the sequence

n = number of terms

d = common difference

Sn = sum of all terms

 

given

a1 = 12

a2 = 16

n = 10

 

d = 16 -12 = 4

@n = 10

a(n) = a1 + (n-1)d

a(10) = 12 + (9)4

a(10) = 48 seats

 

Sn = (n/2) * (a1 + a(10))

Sn = 5* (12 + 48)

Sn = 300 seats

 

Therefore the total number of seats is 300.

 

 

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Mariana [72]

Answer:

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Step-by-step explanation:

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29/sin(115 degrees) = 13/sin(B)

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3 0
3 years ago
What is the cube root of 27a12
satela [25.4K]

Answer:

3a^4

Step-by-step explanation:

Given expression is 27a^{12}.

Now we need to find the cube root of given expression 27a^{12}.

\sqrt[3]{27a^{12}}

=\sqrt[3]{3*3*3a^{4*3}}

=\sqrt[3]{3^3\left(a^4\right)^3}

=\sqrt[3]{\left(3a^4\right)^3}

Since cube root and cube are opposite operations of each other. So they will cancel each other.

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Hence correct choice is the last choice 3a^4.

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3 years ago
Frankie wanted to buy a new TV for his
Mars2501 [29]

Answer:

157.5

Step-by-step explanation:

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hope this helped.

5 0
3 years ago
Find the volume of the solid under the plane 5x + 9y − z = 0 and above the region bounded by y = x and y = x4.
svp [43]
<span>For the plane, we have z = 5x + 9y

For the region, we first find its boundary curves' points of intersection.
x = x^4 ==> x = 0, 1.

Since x > x^4 for y in [0, 1],

The volume of the solid equals

\int\limits^1_0 { \int\limits_{x^4}^x {(5x+9y)} \, dy } \, dx = \int\limits^1_0 {\left[5xy+ \frac{9}{2} y^2\right]_{x^4}^{x}} \, dx  \\  \\ =\int\limits^1_0 {\left[\left(5x(x)+ \frac{9}{2} (x)^2\right)-\left(5x(x^4)+ \frac{9}{2} (x^4)^2\right)\right]} \, dx  \\  \\ =\int\limits^1_0 {\left(5x^2+ \frac{9}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx =\int\limits^1_0 {\left( \frac{19}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx \\  \\ =\left[ \frac{19}{6} x^3- \frac{5}{6} x^6- \frac{1}{2} x^9\right]^1_0

=\frac{19}{6} - \frac{5}{6} - \frac{1}{2} =\bold{ \frac{11}{6} \ cubic \ units}</span>
8 0
3 years ago
Need help with stats!
Brut [27]

Answer:

a) 1,440 ways

b) 59,280 or 64,000

Step-by-step explanation:

a) Aircraft boarding.

8 people, 2 in first class, boarding first, then 8 economy class.

The 2 people in first class board first, but they can board as AB or BA... so 2 ways here.

For the 6 economy class passengers, we have a permutation of 6 out of 6, so 720, as follows:

P(6,6) = \frac{6!}{(6 - 6)!} = 6! = 720

Since the two are independent, we multiply them to have a global number of ways: 2 * 720 = 1,440 different ways for the 8 passengers to board that plane.

b) combination lock.

Here we do have a little problem... the question doesn't specify if the 3 numbers are different numbers of not.  So, we'll calculate both:

Numbers go from 1 to 40 inclusively... so 40 possibilities.

Normally, in a combination lock, the numbers are different, so let's start with that one:

First number: 40 options available

Second number: 39 options available (cannot take the first one again)

Third number: 38 different options (can't take First or Second number again)

Overall, we then have 40 * 39 * 38 = 59,280 different lock combinations.

If we can pick pick the same number twice:

First number: 40 options available

Second number: 40 options available

Third number: 40 options available

Overall 40 * 40 * 40 = 64,000 different lock combinations

8 0
3 years ago
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