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azamat
3 years ago
5

14 1/6-10 11/24= what fraction

Mathematics
2 answers:
spin [16.1K]3 years ago
8 0
The answer is 3 17/24.
DerKrebs [107]3 years ago
3 0
<span>14 1/6-10 11/24
= 14 4/24 - 10 11/24
= 13 28/24 - 10 11/24
= 3 17/24

hope that helps</span>
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brilliants [131]

Answer:

yes your answer is correct

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3 years ago
If x-2 is one of the factors of x^4-kx^2+4, find the value of k.
Andrei [34K]

Step-by-step explanation:

Given expression is x⁴-kx²+4 =0

x−2 is a factor of the given expression.

So, x=2

Substitute in the equation, we get

2⁴-k(2)²+4

⇒16+4k+4=0

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8 0
2 years ago
How to find the area of combined shapes
Elodia [21]
You can split the combined shape into separate shapes and find the area of each of those. once you have done that add all of the areas together and that will get you your combined shapes area.
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3 years ago
2a - 7 - 6a = 9<br> Pls help?
slavikrds [6]

Step-by-step explanation:

2a - 7 - 6a = 9

- 4a - 7 = 9

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3 0
2 years ago
Pretty Pavers company is installing a driveway. Below is a diagram of the driveway they are
prohojiy [21]

Answer:

The most correct option is;

(B) 958.2 ft.²

Step-by-step explanation:

From the question, the dimension of each square = 3 ft.²

Therefore, the length of the sides of the square = √3 ft.

Based on the above dimensions, the dimension of the small semicircle is found by counting the number of square sides ti subtends as follows;

The dimension of the diameter of the small semicircle = 10·√3

Radius of the small semicircle = Diameter/2 = 10·√3/2 = 5·√3

Area of the small semicircle = (π·r²)/2 = (π×(5·√3)²)/2 = 117.81 ft.²

Similarly;

The dimension of the diameter of the large semicircle = 10·√3 + 2 × 6 × √3

∴ The dimension of the diameter of the large semicircle = 22·√3

Radius of the large semicircle = Diameter/2 = 22·√3/2 = 11·√3

Area of the large semicircle = (π·r²)/2 = (π×(11·√3)²)/2 = 570.2 ft.²

Area of rectangle = 11·√3 × 17·√3 = 561

Area, A of large semicircle cutting into the rectangle is found as follows;

A_{(segment \, of \, semicircle)} = \frac{1}{4} \times (\theta - sin\theta) \times r^2

Where:

\theta = 2\times tan^{-1}( \frac{The \, number \, of  \, vertical  \, squrare  \, sides  \ cut  \,  by  \  the  \  large  \,  semicircle}{The \, number \, of  \, horizontal \, squrare  \, sides  \ cut  \,  by  \  the  \  large  \,  semicircle} )

\therefore \theta = 2\times tan^{-1}( \frac{10\cdot \sqrt{3} }{5\cdot \sqrt{3}} ) = 2.214

Hence;

A_{(segment \, of \, semicircle)} = \frac{1}{4} \times (2.214 - sin2.214) \times (11\cdot\sqrt{3} )^2 = 128.3 \, ft^2

Therefore; t

The area covered by the pavers = 561 - 128.3 + 570.2 - 117.81 = 885.19 ft²

Therefor, the most correct option is (B) 958.2 ft.².

4 0
2 years ago
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