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Ilya [14]
3 years ago
12

How to solve 2.1*10^2/8.4*10^5

Mathematics
1 answer:
mariarad [96]3 years ago
7 0
<u> 2.1 × 10²  </u>  =<u>   2.1 × 100 </u>  = <u>210</u>     = 0.00025       
8.4 × 10^5    8.4 × 100000  840000<u>
</u> 
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In circle O, ST is a diameter.
Georgia [21]

Answer:

B)25.0

Step-by-step explanation:

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9. Find (f.g)(a) if f(a) = . and g(a) = 2². Fully simplify your answer.
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6. The dimensions of a living room are 18 ft. by 15 ft. by 8ft. If air weighs 0.08 lb. per cubic foot, the weight of the air in
user100 [1]

The weight of the air in the room is 172.8 lb if the dimensions of a living room are 18 ft. by 15 ft. by 8ft.

<h3>What is a rectangular prism?</h3>

It is defined as the six-faced shape, a type of hexahedron in geometry.

It is a three-dimensional shape. It is also called a cuboid.

It is given that:

The dimensions of a living room are 18 ft. by 15 ft. by 8ft.

The volume of the living room = volume of the cuboid:

V = length×width×height

V = 18×15×8

V = 2160 cubic ft

The weight of the air = 0.08 lb. per cubic foot

The weight of the air in the room = 0.08×2160

The weight of the air in the room = 172.8 lb

Thus, the weight of the air in the room is 172.8 lb if the dimensions of a living room are 18 ft. by 15 ft. by 8ft.

Learn more about the rectangular prism here:

brainly.com/question/21308574

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7 0
2 years ago
Question 1 of 10
andrey2020 [161]

c i think because they dont specify

8 0
3 years ago
Read 2 more answers
An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t
USPshnik [31]
The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

\displaystyle\int_a^bf(x)\,\mathrm dx

Split up the interval [a,b] into n equal subintervals,

[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]

where a=x_0 and b=x_n. Each subinterval has measure (width) \dfrac{a-b}n.

Now denote the left- and right-endpoint approximations by L and R, respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are \{x_0,x_1,\cdots,x_{n-1}\}. Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, \{x_1,x_2,\cdots,x_n\}.

So, you have

L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)
R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)

Now let T denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)

Factoring out \dfrac12 and regrouping the terms, you have

T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)

which is equivalent to

T=\dfrac12\left(L+R)

and is the average of L and R.

So the trapezoidal approximation for your problem should be \dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2
4 0
3 years ago
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