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Nuetrik [128]
3 years ago
10

5/6 divided by 11/13

Mathematics
2 answers:
wolverine [178]3 years ago
6 0

Answer:

65/66

Step-by-step explanation:

5/6 * 13/11 = 65/66

aksik [14]3 years ago
3 0

Answer:

65/66

Step-by-step explanation:

Find the reciprocal of the second number (11/13), which you find by flipping the top and bottom number so it would be 13/11. Now multiply the top two numbers which are 5 and 13, that equals 65. Multiply the bottom numbers, 6 and eleven which is 66. That is how you get 65 over 66.

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Answer:

Dominio de una Raíz cuadrada

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Por lo tanto deberemos estudiar el signo de lo que se encuentra dentro de la raíz; y determinar las zonas positivas y las raíces que son las zona de existencia de la función raíz o dominio de esta función.

Step-by-step explanation:

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3 years ago
Round the number to one decimal place. 67.638
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6 0
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5 0
3 years ago
Find the distance between the complex numbers. z1=9-9i, z2=10-9i
kherson [118]
\bf z_1=\stackrel{a}{9}\stackrel{b}{-9}i\implies (9~,~-9)\qquad \qquad \qquad z_2=\stackrel{a}{10}\stackrel{b}{-9}i\implies (10~,~-9)\\\\
-------------------------------\\\\
~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\

\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~{{ 9}} &,&{{ -9}}~) 
%  (c,d)
&&(~{{ 10}} &,&{{ -9}}~)
\end{array}\quad 
%  distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
d=\sqrt{[10-9]^2+[-9-(-9)]^2}\implies d=\sqrt{(10-9)^2+(-9+9)^2}
\\\\\\
d=\sqrt{1^2+0}\implies d=\sqrt{1}\implies d=1
4 0
3 years ago
the 3rd term of an A.P is 6 and the 16th term is 32 find the common difference and the sum of the first 20 term
Digiron [165]

Answer:

2 and 420

Step-by-step explanation:

The n th term of an AP is

a_{n} = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Given a₃ = 6 and a₁₆ = 32 , then

a₁ + 2d = 6 → (1)

a₁ + 15d = 32 → (2)

Subtract (1) from (2) term by term

13d = 26 ( divide both sides by 13 )

d = 2

Substitute d = 2 into (1) for corresponding value of a₁

a₁ + 2(2) = 6

a₁ + 4 = 6 ( subtract 4 from both sides )

a₁ = 2

The sum of n terms of an AP is

S_{n} = \frac{n}{2} [ 2a₁ + (n - 1)d ] , thus

S_{20} = \frac{20}{2} [(2 × 2) + (19 × 2) ]

     = 10(4 + 38) = 10 × 42 = 420

6 0
4 years ago
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