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3 years ago

Write 3/50 as a decimal and as a percent i need help

Mathematics

1 answer:

Alex17521 [72]3 years ago

7 0

Answer:

0.06 and to change that to a percentage all you do is move the decimal point 2 places to the right so the answer would be <u>6</u><u> </u>percent

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Are the following lines parallel perpendicular or neither y=-5/3x - 5 and y=4/3x - 5

katrin [286]

Answer:

neither

Step-by-step explanation:

For lines to be parallels the slopes must be equal, for the lines to be perpendicular the lines have to be inverse reciprocals meaning if you have a slope of 1/2X then the perpendicular line must be -2X

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2 years ago

Which transformations will map octagon PQRSTVWZ onto itself?

Usimov [2.4K]

Answer:

Rotation, Reflection and Translation.

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2 years ago

Which figure is not a parallelogram?

lara31 [8.8K]

Answer:

the second one

Step-by-step explanation:

it does not say that both sides are parallel

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3 years ago

Read 2 more answers

1) State the area of the given trapezoid.

Ber [7]

Answer:

The answer would be 26

Step-by-step explanation:

First I divided the shape to make a rectangle and a right angle triangle, then i calculated the shape of the rectangle which came out to be 20. After that I calculated the area of the right angle triangle which is 6. So The answer would be 6 + 20 = 26

5 0

2 years ago

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

11 months ago