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Bogdan [553]
3 years ago
12

Prevent as a decimal 21%

Mathematics
1 answer:
zalisa [80]3 years ago
7 0

Answer:

0.21

Step-by-step explanation:

21 / 100 =0.21

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10 five digit numbers can be formed using only odd digits
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A square with side length C has an area of 81 square centimeters. The following equation shows the area of the square.
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3 years ago
If f(x) = 2x + 1 and g(x) = x^2 find g(f(3))<br> 49<br> 19<br> 7<br> 13
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Answer:

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Put the numbers in place of the variables and do the arithmetic.

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3 0
4 years ago
A rectangular field is 75 yards wide and 105 yards long
nika2105 [10]

Answer:

width = 72 yards

length = 108 yards

Step-by-step explanation:

Given:

  • Width = 75 yards
  • Length = 105 yards

<u>Area of the field</u> with the given values:

\begin{aligned}\textsf{Area of a rectangle}&=\sf width \times length\\& = \sf 75 \times 105\\& = \sf 7875\:\: yd^2\end{aligned}

To maintain the <u>same perimeter</u>, but <u>change the area</u>, either:

  • decrease the width and increase the length by the same amount, or
  • increase the width and decrease the length by the same amount.

In geometry, length pertains to the <u>longest side</u> of the rectangle while width is the <u>shorter side</u>.  Therefore, we should choose:

  • decrease the <u>width</u> and increase the <u>length</u> by the <u>same amount</u>.

<u>Define the variables</u>:

  • Let x = the amount by which to decrease/increase the width and length.

Therefore:

\implies \sf width \times length < 7875\:\:yd^2

\implies (75-x)(105+x) < 7875

Solve the inequality:

\begin{aligned}(75-x)(105+x) & < 7875\\7875-30x-x^2 & < 7875\\-x^2-30x & < 0\\-x(x+30) & < 0\\x(x+30) & > 0\\\implies x & > 0 \:\: \textsf{ or }\:\:x < - 30\end{aligned}

Therefore, as distance is positive only and the maximum width is 75 yd (since we are subtracting from the original width):

\begin{cases}\textsf{width} = 75 - x\\\textsf{length} = 105 + x\end{cases}

\textsf{where } 0 < x < 75

Therefore, to find the width and length of another rectangular field that has the same perimeter but a smaller area than the first field, simply substitute a value of x from the restricted interval into the found expressions for width and length:

<u>Example 1</u>:

  • Let x = 3

⇒ Width = 75 - 3 = 72 yd

⇒ Length = 105 + 3 = 108 yd

⇒ Perimeter = 2(72 + 108) = 360 yd

⇒ Area = 72 × 108 = 7776 yd²

<u>Example 2</u>:

  • Let x = 74

⇒ Width = 75 - 74 = 1 yd

⇒ Length = 105 + 74 = 179 yd

⇒ Perimeter = 2(1 + 179) = 360 yd

⇒ Area = 1 × 179 = 179 yd²

4 0
2 years ago
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