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8090 [49]
3 years ago
11

Square root of 64 + cube root of 64 is​

Mathematics
2 answers:
seropon [69]3 years ago
8 0

Answer:

12

Step-by-step explanation:

\sqrt{64} + \sqrt[3]{64}

= 8 + 4

= 12

ValentinkaMS [17]3 years ago
3 0

Square root of 64 = \sf \sqrt{64}  =  \sqrt{8 \times 8}  =  \boxed{\bf 8}

Cube root of 64 = \sf \sqrt[3]{64}  =   \sqrt[3]{4 \times 4 \times 4}  =  \boxed{ \bf4}

So,

Square root of 64 + Cube root of 64

=  \sf \sqrt{64}  +  \sqrt[3]{64}  \\ \sf  = 8 + 4 \\  =  \boxed{ \bf12}

<u>1</u><u>2</u><u> </u>is the answer of the question asked.

____________

Hope it helps.

RainbowSalt2222

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Using the binomial theorem , obtain the expansion of :
andrezito [222]

Answer:

see explanation

Step-by-step explanation:

Expand both factors and collect like term

Using Pascal' triangle with n = 6 to obtain the coefficients

1  6  15  20  15  6  1

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Increasing powers of 3x from (3x)^{0} to (3x)^{6}

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= 1.1^{6}(3x)^{0} + 6.1^{5}(3x)^{1} + 15.1^{4}(3x)^{2} + 20.1^{3}(3x)^{3} + 15.1²(3x)^{4} + 6.1^{1}(3x)^{5} + 1.1^{0}(3x)^{6}

= 1 + 18x + 135x² + 540x³ + 1215x^{4} + 1458x^{5} + 729x^{6}

--------------------------------------------------------------------------------------

(1-3x)^{6}

= 1.1^{6}(-3x)^{0} + 6.1^{5}(-3x)^{1} + 15.1^{4}(-3x)^{2} + 20.1^{3}(-3x)^{3} + 15.1²(-3x)^{4} + 6.1^{1}(-3x)^{5} + 1.1^{0}(-3x)^{6}

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----------------------------------------------------------------------------------

Collecting like terms from both expressions

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----------------------------------------------------

(2)

Using Pascal's triangle with n = 5

1  5  10  10  5  1

Decreasing powers of 1 from 1^{5} to 1^{0}

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(1+2x)^{5}

= 1.1^{5}(2x)^{0} + 5.1^{4}(2x)^{1} + 10.1^{3}(2x)^{2} + 10.1^{2}(2x)^{3} + 5.1^{1}(2x)^{4}+ 1.1^{0}(2x)^{5}

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8 0
3 years ago
Peyton is collecting data on the lunch habits of students at Hyattsville Middle School. In one class, 14 students bring lunch fr
mart [117]

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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