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3 years ago

Find square root of 7921

Mathematics

2 answers:

V125BC [204]3 years ago

8 0

Answer: 89

89 times 89 is 7921. Since 7921 is a 4 digit perfect square, it has 89 as its root. Square root of 7921=89

Bas_tet [7]3 years ago

6 0

Answer:89

Step-by-step explanation:

√7921

89|7921

89|89

√89×89=89

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7.98 is 1/10 of what number?

Lyrx [107]

Answer:

79.8

Step-by-step explanation:

7.98*10=

79.8

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(1/10)*79.8=

7.98

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3 years ago

The area of a rectangle is 42 milimeter .The length is 7 milimeters. What is the perimter of rectangel?

Leni [432]

Answer:

6 millimeters

Step-by-step explanation:

A = L × W

A ÷ L = W

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3 years ago

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Expand and simplify (2x+1)(x+3)

Len [333]

Answer:

2x^2+7x+3

Step-by-step explanation:

(2x+1)(x+3)

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3 years ago

Can anyone help me about question 2

madreJ [45]

9514 1404 393

Answer:

3(4/3)^2543

Step-by-step explanation:

Using logarithms base 2, we have ...

$(a_n)^{\log{a_n}}=(a_{n+1})^{\log{a_{n-1}}}\\\\(\log{a_n})(\log{a_n}) = (\log{a_{n-1}})(\log{a_{n+1}}) \\\\ \dfrac{\log{a_{n+1}}}{\log{a_{n}}}=\dfrac{\log{a_n}}{\log{a_{n-1}}}=\dots\dfrac{\log_2{16}}{\log_2{8}}=\dfrac{4}{3}$

That is, the ratio of each term to the previous is a constant equal to 4/3. This is the definition of a geometric sequence. This sequence has first term 3 and common ratio 4/3, so the general term is ...

$\log_2 a_n=3\left(\dfrac{4}{3}\right)^{n-1}$

and the 2544th term is ...

$\boxed{\log_2{a_{2544}}=3\left(\dfrac{4}{3}\right)^{2543}}$

8 0

3 years ago

Find square root of 7921​

Find square root of 7921