(2x+5+9) (3x+6)[tex](2x + 5 + 9)(3x + 6)[/tex]Apply BODMAS to solve thisNo, spam answers please ​

  • matematika

    Subject:

    Math

  • Author:

    chen

  • Created:

    1 year ago

Answers 2

1

Add the numbers

(2x+{5}+{9})(3x+6)

(2x+{14})(3x+6)

2

Distribute

{(2x+14)(3x+6)}

{2x(3x+6)+14(3x+6)}

3

Distribute

{2x(3x+6)}+14(3x+6)

{6x^{2}+12x}+14(3x+6)

4

Distribute

6x^{2}+12x+{14(3x+6)}

6x^{2}+12x+{42x+84}

5

Combine like terms

6x^{2}+{12x}+{42x}+84

6x^{2}+{54x}+84

Solution

6x^{2}+54x+84

[tex](2x - 5 + 9)(3x + 6)[/tex]

[tex](2x - 14)(3x + 6)[/tex]

[tex]2x(3x + 6) + 14(3x + 6)[/tex]

[tex]6x^{2} + 12x + 14(3x + 6)[/tex]

[tex]6x^{2} + 54x + 84[/tex]

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Answer:

Correct option is B)

Reductive amination is a way by which ammonium ions is assimilated to form amino acids in plants. In this reaction, in the presence of enzyme glutamate dehydrogenase, ammonium ion directly combines with alpha-ketoglutaric acid, to form glutamic acid (amino acid) and for this to happen, a reduced coenzyme (NADPH) is required.

So the correct answer is 'Reductive amination'.

When 1 moe of ice melts at 0oC and at constant pressure of 1 atm, 1440 calories of heat are absorbed by the system. The molar volume of ice and water are 0.0196 and 0.0180 litre respectively. Which of the following is correct? a.ΔU≈ΔH b.ΔU=1548 cal c.ΔU=1056 cal d.None of the above

Answer:

Thus, ΔH≈ΔU=1440 calories. Q. When 1 moe of ice melts at 0oC and at constant pressure of 1 atm, 1440 calories of heat are absorbed by the system. The molar volume of ice and water are 0.0196 and 0.0180

Which type of junction faciliitates the cells to communicate with each other by connecting the cytoplasm of adjoing cells? a.Tight junction b.Adhering junction c.Gap junction d.Both a & c

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A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with their magnetic moments parallel to their respective axes. But the magnetic moment of hoop is twice that of solid cylinder. They are then placed in a uniform magnetic field in such a manner that their magnetic moments make a small angle with the field. If the oscillation periods of hoop and cylinder are Th and Tc respectively, then a.Th=Tc b.Th=32 Tc c.Th=2 Tc d.Th=12 Tc

Answer:

b.Th=32 Tc

Explanation:

A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with their magnetic moments parallel to their respective axes. But the magnetic moment of hoop is twice that of solid cylinder. They are then placed in a uniform magnetic field in such a manner that their magnetic moments make a small angle with the field. If the oscillation periods of hoop and cylinder are Th and Tc respectively, then

[tex]Th=2π \sqrt{ \frac{

mR {}^{2} }{(2μ)B} }

[/tex]

[tex]TC =2 \pi \sqrt{ \frac{ \frac{1}{2}m {R}^{2} }{μB} } [/tex]

answer a is correct Th = Tc

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