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STatiana [176]
3 years ago
12

Evaluate c-2 when c=7

Mathematics
2 answers:
Luba_88 [7]3 years ago
8 0
5 is the answer. Hope this helps.
zhannawk [14.2K]3 years ago
8 0

Answer:

The value of the expression c-2 when c=7 is 5.

Step-by-step explanation:

Consider the provided expression.

We need to find the value of expression for c=7.

Substitute the value of c=7 in the provided expression and simplify as shown.

c-2

7-2

Subtract the numbers as shown:

5

Hence, the value of the expression c-2 when c=7 is 5.

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Please help SO CONFUSED
gulaghasi [49]

Answer:

SOH-CAH-TOA

h = \sqrt{ 9^{2} +5^{2} }

~~~~~~~~~~

H= \sqrt{106}

O= 5

A= 9

~~~~~~~~~~~~~~~~

Sin = \frac{5}{\sqrt{106 }  }

Cos= \frac{9}{\sqrt{106 }  }

Tan = \frac{5}{9  }

Step-by-step explanation:

5 0
3 years ago
Find the distance between the pair of points<br><br>(-13,11) and (-6,5)​
Alex_Xolod [135]
Pick me as the brainliest answe please

8 0
3 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
3 years ago
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Advocard [28]

Answer:

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Step-by-step explanation:

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