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Annette [7]
3 years ago
5

Find the length of rhombus whose diogonals are length 12 cm and 16 cm​

Mathematics
1 answer:
grin007 [14]3 years ago
7 0

In ΔAOD  (AD)  =(AO)

2+(OD)

2   =36+64    AD

2 =100

  AD= 100

=10cm

 Rhombus has all sides equal

AB = AD = BC = CD = 10 cm

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123 + 12 + 1 + 1 + 1 =
PilotLPTM [1.2K]

Step-by-step explanation:

123 +

 12

______

   5              (3+2)

 3                (2+1)

1                    (1)

so

123 + 12 = 135

1 + 1 + 1 = 3

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comment answer

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2 years ago
What are the values of the digits in 8,040,930?
kkurt [141]

Answer:

the second option

Step-by-step explanation:

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3 years ago
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3 years ago
Evaluate the indicated limit algebraically. Change the form of the function where necessary. Please write clearly with descripti
alukav5142 [94]

Answer:

\displaystyle \lim_{x\to \infty}\frac{3x^2+4.5}{x^2-1.5}=3

Step-by-step explanation:

We want to evaluate the limit:

\displaystyle \lim_{x\to \infty}\frac{3x^2+4.5}{x^2-1.5}

To do so, we can divide everything by <em>x</em>². So:

=\displaystyle \lim_{x\to \infty}\frac{3+4.5/x^2}{1-1.5/x^2}

Now, we can apply direct substitution:

\Rightarrow \displaystyle \frac{3+4.5/(\infty)^2}{1-1.5/(\infty)^2}

Any constant value over infinity tends towards 0. Therefore:

\displaystyle =\frac{3+0}{1+0}=\frac{3}{1}=3

Hence:

\displaystyle \lim_{x\to \infty}\frac{3x^2+4.5}{x^2-1.5}=3

Alternatively, we can simply consider the biggest term of the numerator and the denominator. The term with the strongest influence in the numerator is 3<em>x</em>²<em>, </em>and in the denominator it is <em>x</em>². So:

\displaystyle \Rightarrow \lim_{x\to\infty}\frac{3x^2}{x^2}

Simplify:

\displaystyle =\lim_{x\to\infty}3=3

The limit of a constant is simply the constant.

We acquire the same answer.

6 0
3 years ago
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