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Vika [28.1K]
3 years ago
6

A rectangle has a length of 12 meters and a width of 400 centimeters. What is the perimeter,in cm, of the rectangle?

Mathematics
1 answer:
aniked [119]3 years ago
7 0

Step-by-step explanation:

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The right triangle on the right is a scaled
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Answer:

The scale factor is 3.5

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PLEASE PLEASE PLEASEE PLEASEEEEESE HELPME!?!?!? IM SO BAD AT MATH PLEASE HELP!!!
Fiesta28 [93]
For this case we have the following function:
 f (x) = 0.01 * (2) ^ x

 By definition, the average rate of change is given by:
 AVR =  \frac{f(x2) - f(x1)}{x2 - x1}
 We evaluate the function for the given values:
 For x = 7:
 f (7) = 0.01 * (2) ^ 7

f (7) = 1.28
 For x = 14:
 f (14) = 0.01 * (2) ^ {14}  f (14) = 163.84
 Then, replacing values we have:
 AVR = \frac{163.84 - 1.28}{14 - 7}
 AVR = 23.22
 Answer:
 
the average rate of change from x = 7 to x = 14 is:
 
a. 23.22 
7 0
3 years ago
Find an odd natural numbers x such that LCM (x, 40) = 1400​
vesna_86 [32]

Answer:

Let x be odd such that LCM {x,40} = 1400 .

Since 1400 = 23×52×7 , then

x ∈ {5m×7n∣(m,n) ∈ {0,1,2}×{0,1}} .

By testing these values, we find that x = 175 .

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You go to a store to pick up milk and cereal. You see that the milk costs $4
PolarNik [594]
Use the estimate but that should be more than the exact price
4 0
3 years ago
Read 2 more answers
You have D dollars to buy fence to enclose a rectangular plot of land (see figure at right). The fence for the top and bottom co
alex41 [277]

The perimeter of the rectangular plot of land is given by the expression below

P=2x+2y

On the other hand, since the available money to buy fence is D dollars,

\begin{gathered} D=4(2x)+3(2y) \\ \Rightarrow D=8x+6y \\ D\rightarrow\text{ constant} \end{gathered}

Furthermore, the area of the enclosed land is given by

A=xy

Solving the second equation for x,

\begin{gathered} D=8x+6y \\ \Rightarrow x=\frac{D-6y}{8} \end{gathered}

Substituting into the equation for the area,

\begin{gathered} A=(\frac{D-6y}{8})y \\ \Rightarrow A=\frac{D}{8}y-\frac{3}{4}y^2 \end{gathered}

To find the maximum possible area, solve A'(y)=0, as shown below

\begin{gathered} A^{\prime}(y)=0 \\ \Rightarrow\frac{D}{8}-\frac{3}{2}y=0 \\ \Rightarrow\frac{3}{2}y=\frac{D}{8} \\ \Rightarrow y=\frac{D}{12} \end{gathered}

Therefore, the corresponding value of x is

\begin{gathered} y=\frac{D}{12} \\ \Rightarrow x=\frac{D-6(\frac{D}{12})}{8}=\frac{D-\frac{D}{2}}{8}=\frac{D}{16} \end{gathered}<h2>Thus, the dimensions of the fence that maximize the area are x=D/16 and y=D/12.</h2><h2>As for the used money,</h2>\begin{gathered} top,bottom:\frac{8D}{16}=\frac{D}{2} \\ Sides:\frac{6D}{12}=\frac{D}{2} \end{gathered}<h2>Half the money was used for the top and the bottom, while the other half was used for the sides.</h2>

7 0
9 months ago
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