Answer: The approximate area of the circle is: 3,629.84 cm² .
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Area of a circle, "A" : A = π * r² ;

in which: A = Area of circle ;
π ≈ 3.14 ;
 r = radius;

radius of a circle, "r" = d / 2 ; in which: "d = diameter" ;

Given "d = 64 cm " ;

r = d / 2 = 64 cm / 2 = 32 cm ;
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So; A = π * r² ;

A ≈ (3.14) * (32 cm)² ;

A ≈ (3.14) * (1156 cm²) ;
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A ≈ 3,629 .84 cm² .
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3 0

2 years ago

DF bisects EDG Find FG. The diagram is not to scale. A.15 B.14 C.19 D. 28

Ahat [919]

Answer:

○ b. $14$

Step-by-step explanation:

Since DF is a segment bisector, set both expressions equal to each other:

3n - 4 = n + 8

-3n - 3n

__________

−4 = −2n + 8

-8 - 8

_________

−12 = −2n

__ ___

−2 −2

$6 = n$ [Plug this back into both expressions above to get the value of 14]; $14 = FG; 14 = EF$

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6 0

3 years ago

Given that 1 x2 dx 0 = 1 3 , use this fact and the properties of integrals to evaluate 1 (4 − 6x2) dx. 0

Debora [2.8K]

So, the definite integral $\int\limits^1_0 {(4 - 6x^{2} )} \, dx= - 74$

Given that

$\int\limits^1_0 {x^{2} } \, dx = 13$

We find

$\int\limits^1_0 {(4 - 6x^{2} )} \, dx$

<h3>Definite integrals </h3>

Definite integrals are integral values that are obtained by integrating a function between two values.

So, $Integral \int\limits^1_0 {(4 - 6x^{2} )} \, dx$

So, $\int\limits^1_0 {(4 - 6x^{2} )} \, dx = \int\limits^1_0 {4} \, dx - \int\limits^1_0 {6x^{2} } \, dx \\= 4[x]^{1}_{0} - \int\limits^1_0 {6x^{2} } \, dx \\= 4[x]^{1}_{0} - 6\int\limits^1_0 {x^{2} } \, dx \\= 4[1 - 0] - 6\int\limits^1_0 {x^{2} } \, dx\\= 4[1] - 6\int\limits^1_0 {x^{2} } \, dx\\= 4 - 6\int\limits^1_0 {x^{2} } \, dx$

Since

$\int\limits^1_0 {x^{2} } \, dx = 13$ ,

Substituting this into the equation the equation, we have

$\int\limits^1_0 {(4 - 6x^{2} )} \, dx = 4 - 6\int\limits^1_0 {x^{2} } \, dx\\= 4 - 6 X 13 \\= 4 - 78\\= -74$

So, $\int\limits^1_0 {(4 - 6x^{2} )} \, dx= - 74$

Learn more about definite integrals here:

brainly.com/question/17074932

4 0

2 years ago