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SashulF [63]
3 years ago
15

Si el lado de un cuadrado mide 10cm cuanto mide él área del círculo que se encuentra adentro de ese cuadrado

Mathematics
1 answer:
Alex17521 [72]3 years ago
8 0

Answer:

El área del círculo que se encuentra en el cuadrado es de 78.5cm²

Step-by-step explanation:

Para resolver este ejercicio tenemos que pensar que un cuadrado tiene sus 4 lados iguales, por lo que todos sus lados medirán 10cm.

Ahora nos fijamos que necesitamos saber para calcular el área de un circulo

a = área

r = radio

π = 3.14

a = π * r²

como podemos ver no sabemos el valor del radio

como el circulo toca con los 4 lados del cuadrado sabemos que su radio sera la distancia del centro del cuadrado a cualquiera de los lados.

Entonces tenemos que dividir un lado por 2

10cm/2 = 5cm

El radio del circulo sera 5cm

Ahora que tenemos todos los datos podemos calcular el valor del área

a = 3.14 * (5cm)²

a = 3.14 * 25cm²

a = 78.5cm²

El área del círculo que se encuentra en el cuadrado es de 78.5cm²

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amm1812

Answer:

11 dimes and 16 quarters

Step-by-step explanation:

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3 years ago
Explain why the two figures below are similar. Use complete sentences and provide evidence to support your explanation. (10 poin
alekssr [168]

Answer:

<em>The two figures are similar as  the ratio of corresponding distances and slope of corresponding lines are EQUAL.</em>

Step-by-step explanation:

Noting the coordinates of points,

C(-1,6), D(3,4) , E(2,3) , F(3,0), G(1,-2) , A(0,1) , B(-2,2)

J(4.5,3.5), K(6.5,2.5), L(6,2), M(6.5,0.5) , N(5.5,-0.5) , H(5,1) , I(4,1.5)

using the distance formula = \sqrt{(x-x1)^{2} + (y-y1)^{2} }

we find that,

CD=\sqrt{20} and JK=\sqrt{5}

thus, ratio of CD to JK is 2.

Similarly find ratio of DE to KL and so on, the ratio remains same.

Further one can even comment on the slopes of corresponding lines been equal.

Thus the figures turn out to be similiar to each other

7 0
3 years ago
Need to Simplify 6^5/6^3
katrin [286]

Answer:

6^2

Step-by-step explanation:

We know a^b / a^c = a^(b-c)

6^5 / 6^3 = 6^(5-3)  = 6^2

3 0
3 years ago
What is the nth term rule of the quadratic sequence below?<br> -5, -3,3, 13, 27, 45, 67, ...
Svetllana [295]

Answer:

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so its in increment of 4

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Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Find the derivative of the function f(x) = (x3 - 2x + 1)(x – 3) using the product rule.
julsineya [31]

Answer:

Step-by-step explanation:

Hello, first, let's use the product rule.

Derivative of uv is u'v + u v', so it gives:

f(x)=(x^3-2x+1)(x-3)=u(x) \cdot v(x)\\\\f'(x)=u'(x)v(x)+u(x)v'(x)\\\\ \text{ **** } u(x)=x^3-2x+1 \ \ \ so \ \ \ u'(x)=3x^2-2\\\\\text{ **** } v(x)=x-3 \ \ \ so \ \ \ v'(x)=1\\\\f'(x)=(3x^2-2)(x-3)+(x^3-2x+1)(1)\\\\f'(x)=3x^3-9x^2-2x+6 + x^3-2x+1\\\\\boxed{f'(x)=4x^3-9x^2-4x+7}

Now, we distribute the expression of f(x) and find the derivative afterwards.

f(x)=(x^3-2x+1)(x-3)\\\\=x^4-2x^2+x-3x^3+6x-4\\\\=x^4-3x^3-2x^2+7x-4 \ \ \ so\\ \\\boxed{f'(x)=4x^3-9x^2-4x+7}

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

6 0
3 years ago
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