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3 years ago

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

Whitepunk [10]

Answer:

The scale factor is 3.5

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2 years ago

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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3 years ago

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