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Hoochie [10]
3 years ago
6

Solve x and y x/5+y/6=12

Mathematics
1 answer:
Arlecino [84]3 years ago
4 0

Answer:

2435

Step-by-step explanation:

add your numbers

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Answer:

0.6 over 0.6. Or just 1.

Step-by-step explanation:

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2 years ago
50:10 in simplest form
Reil [10]

Answer:

5/1 or 5:1

Step-by-step explanation:

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A circle is drawn within a square as shown.
Vinil7 [7]
<h3>Area of shaded region is 21.5 square centimeter</h3>

<em><u>Solution:</u></em>

The area of the shaded region is equal to the area of the square minus the area of the circle

<em><u> Find the area of the square</u></em>

Area\ of\ square = a^2

Where, "a" is the length of each side

From given figure in question,

a = 10 cm

Area\ of\ square = 10^2\\\\Area\ of\ square = 100\ cm^2

<em><u>Find the area of circle </u></em>

The area of the circle is given as:

Area\ of\ circle = \pi r^2

Where, "r" is the radius of circle

Diameter = 10 cm

r = \frac{diameter}{2}\\\\r = \frac{10}{2}\\\\r = 5

Thus,

Area\ of\ circle = 3.14 \times 5^2\\\\Area\ of\ circle = 3.14 \times 25\\\\Area\ of\ circle = 78.5\ cm^2

<em><u>Find the area of the shaded region</u></em>

Area of shaded Region = Area of Square - Area of Circle

Area of shaded Region = 100 - 78.5 = 21.5

Thus area of shaded region is 21.5 square centimeter

7 0
2 years ago
Please help me ASAP
podryga [215]

Answer:ok

Step-by-step explanation:

8 0
3 years ago
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Please help .god bless you
zhenek [66]

Answer:Area of the lawn is 1725 ft^2

Step-by-step explanation:

The yard is in the shape of a trapezoid. The area of the lawn can be determined by finding the area of the trapezoid. The formula for determining the area of a trapezoid is expressed as

Area of trapezoid =

1/2(a + b)h

Where

a is the length of one of the parallel sides of the trapezoid

b is the length of the other parallel side of the trapezoid.

h is the perpendicular height of the the trapezoid.

From the diagram,

a = 50 feet

b = 65 feet

h = 30 feet

Area of the lawn = 1/2(50 + 65)× 30

= 1/2 × 115 × 30 = 1725 ft^2

7 0
2 years ago
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