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3 years ago

Can you write 2x(x-1) as a product and as a sum please? Thank you

Mathematics

2 answers:

kotykmax [81]3 years ago

5 0

Answer: yes you can actually.

Step-by-step explanation:

choli [55]3 years ago

5 0

Answer: The product is 2x^2-2x, and the sum is 3x-1

Step-by-step explanation: Assuming you mean 2x(x-1) and 2x+(x-1) it would be 2x^2-2x for the product, and for the sum it would be 3x-1

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melamori03 [73]

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3 years ago

Read 2 more answers

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horrorfan [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

$g(h(x)) = \sqrt[4]{x^2 + x +5} + 3$

$h(g(x)) = \sqrt{x} + 7\sqrt[4]{x} + 17$

$h(h(x)) = [x^2 + x + 5 ]^2 + x^2 + x + 10$

Step-by-step explanation:

From the question we are told that

$h(x) = x^2 + x + 5$

and

$g(x) = \sqrt[4]{x} + 3$

Considering first question

Now we are told g(h(x))

i.e

$g(h(x)) = [x^2 + x + 5 ]^{\frac{1}{4} } + 3$

=> $g(h(x)) = \sqrt[4]{x^2 + x +5} + 3$

Considering second question

Now we are told h(g(x))

i.e

$h(g(x)) = [x^{\frac{1}{4} } + 3]^2 + x^{\frac{1}{4} } + 3 + 5$

=> $h(g(x)) = x^{\frac{1}{2} } + 6x^{\frac{1}{4} } + 9+ x^{\frac{1}{4}} + 8$

=> $h(g(x)) = x^{\frac{1}{2}} + 7x^{\frac{1}{4}} + 17$

=> $h(g(x)) = \sqrt{x} + 7\sqrt[4]{x} + 17$

Considering third question

$h(h(x))= [x^2 + x + 5]^2 + [x^2 + x + 5 ] + 5$

=> $h(h(x)) = [x^2 + x + 5 ]^2 + x^2 + x + 10$

8 0

4 years ago

Which inequality is true? A. 12pi > 4 B. 8pi > 24 C. 6 – pi > 3 D. pi + 3 < 6

Sever21 [200]

The answer you should have is c That should be able To help you My friend Message if I’m wrong and I’ll fix it for you

6 0

3 years ago

3.4 m = cm help plezzzzz

AnnZ [28]

Answer:

thee answer shall be 3.4m = 340cm

6 0

3 years ago

2 2

ExtremeBDS [4]

Answer:its 5

Step-by-step explanation:

4 0

3 years ago