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BigorU [14]
3 years ago
12

A large auditorium has stadium with stadium seating. There are 8 seats in the first row, 10 seats in the second row, 12 seats in

the third row and 300 seats in the last row. How many rows of seats are in the auditorium
Mathematics
1 answer:
PIT_PIT [208]3 years ago
4 0
There are 4 rows, and 338 seats total.
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NeTakaya

Answer:

the answer is the last one

Step-by-step explanation:

3 0
2 years ago
Pls, can you help me? thx.<br><img src="https://tex.z-dn.net/?f=%20%5Ccos%28x%20%2B%20%5Cfrac%7B%5Cpi%7D%7B3%7D%29%20%5Cgeqslant
Veseljchak [2.6K]

For x between -\pi and \pi, we have

  • \cos x=\frac{\sqrt2}2=\frac1{\sqrt2} when x=\pm\frac\pi4;
  • \cos x=1 for x=0; and
  • \cos x=0 for x=\pm\frac\pi2

\cos x is continuous over its domain, so the intermediate value theorem tells us that

\cos x\ge\frac{\sqrt2}2

is true for -\frac\pi4\le x\le\frac\pi4.

For all x, we take into account that \cos x is 2\pi-periodic, so the above inequality can be expanded to

-\dfrac\pi4\le x+2n\pi\le\dfrac\pi4

where n is any integer. Equivalently,

-\dfrac\pi4-2n\pi\le x\le\dfrac\pi4-2n\pi

To get the corresponding solution set for

\cos\left(x+\dfrac\pi3\right)\ge\dfrac{\sqrt2}2

simply replace x with x+\frac\pi3:

-\dfrac\pi4-2n\pi\le x+\dfrac\pi3\le\dfrac\pi4-2n\pi

\implies\boxed{-\dfrac{7\pi}{12}-2n\pi\le x\le-\dfrac\pi{12}-2n\pi}

7 0
3 years ago
. Consumer Awareness In 1986, the average cost of a new midsize four-door sedan was $9000. In 1991, the average cost was $12,000
eduard

<u>Answer-</u>

\boxed{\boxed{y=9000(1.0592)^x}}

<em>And cost of a car in the year 2000 will be </em><em>$20134.18</em>

<u>Solution-</u>

Let's assume,

x = number of years after 1986

y = average cost of sedan in dollar

The exponential model that will model the scenario will be in the form of,

y=ab^x

where a and b are constants

As given that, in 1986 average cost of sedan was $9000 and in 1991 average cost of sedan was $12000

So, the points (0, 9000) and (5, 12000) will satisfy or lie on the exponential curve.

Putting (0, 9000) in the equation,

\Rightarrow 9000=ab^0

\Rightarrow a\times 1=9000

\Rightarrow a=9000

Now, the equation becomes y=9000b^x

Putting (5, 12000) in this equation,

\Rightarrow 12000=9000b^5

\Rightarrow b^5=\dfrac{12000}{9000}

\Rightarrow b=\sqrt[5]{\dfrac{4}{3}}

\Rightarrow b=1.0592

Putting the values,

y=9000(1.0592)^x

As we have to calculate the cost of sedan in 2000, so putting x=14(as 2000-1986=14),

y=9000(1.0592)^{14}=20134.18

Therefore, cost of a car in the year 2000 will be $20134.18

8 0
3 years ago
James molded a fraction by shadding the parts of the circle as shown
icang [17]

the answer that is equal to 1/4 is 2/8 3/12 and 4/16

4 0
3 years ago
Can someone help me with these questions I would really appreciate it
Eva8 [605]

Answer:

a. c^2-b^2=a^2

b. 14

7 0
3 years ago
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