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JulsSmile [24]
3 years ago
14

What is the following sum? 5(3√x)+9(3√x)

Mathematics
2 answers:
Liula [17]3 years ago
4 0

Answer:

Option (c) is correct.

5(\sqrt[3]{x})+9(\sqrt[3]{x} )=14\sqrt[3]{x}

Step-by-step explanation:

Given : expression 5(\sqrt[3]{x})+9(\sqrt[3]{x} )

We have to choose the correct option from the given options to the sum of the given expression.

Consider the given expression 5(\sqrt[3]{x})+9(\sqrt[3]{x} )

Taking \sqrt[3]{x} common from both terms, we have,

(\sqrt[3]{x})(5+9)=14\sqrt[3]{x}

Thus, 5(\sqrt[3]{x})+9(\sqrt[3]{x} )=14\sqrt[3]{x}

charle [14.2K]3 years ago
3 0
5(3√x)+9(3√x)
=3√x(5+9)
=14(3√x)

answer is C. third choice 14(3√x)
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A) Write the sequence of natural numbers which are multiplied by 3 ?
Volgvan

Answer:

a) 3, 6, 9, 12, 15,...,3\cdot n, b) 4, 7, 10, 13, 16,...,3\cdot n +1, c) Both sequences are arithmetic.

Step-by-step explanation:

a) The sequence of natural numbers which are multiplied by 3 are represented by the function f(n) = 3\cdot n, n\in \mathbb{N}. Let see the first five elements of the sequence: 3, 6, 9, 12, 15,...

b) The sequence of natural numbers which are multiplied by 3 and added to 1 is represented by the function f(n) = 3\cdot n + 1, n\in \mathbb{N}. Let see the first five elements of the sequence: 4, 7, 10, 13, 16,...

c) Both sequences since differences between consecutive elements is constant. Let prove this statement:

(i) f(n) = 3\cdot n

\Delta f = f(n+1) -f(n)

\Delta f = 3\cdot (n+1) -3\cdot n

\Delta f = 3

(ii) f(n) = 3\cdot n +1

\Delta f = f(n+1)-f(n)

\Delta f = [3\cdot (n+1)+1]-(3\cdot n+1)

\Delta f = 3

Both sequences are arithmetic.

8 0
3 years ago
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