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3 years ago

What is the fourth term of the expansion of the binomial (2x + 5)^5? Hence find the value of (2.02)5 correct 3D.P

Mathematics

1 answer:

vazorg [7]3 years ago

7 0

Answer:

5000x^2 is the 4th term

Step-by-step explanation:

ok, 1 bionimial theorem coming right up

for a binomial expansion of (a+b)^n

the kth term is

means

n is 5

4th term

4-1=3

and a=2x

b=5

(10)(4x^2)(125)

5000x^2 is the 4th term

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Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

etc.

3 0

3 years ago

How do you decide which variable to isolate in order to use the substitution method to solve a system of linear equation?

Kazeer [188]

It doesn’t really matter which variable you isolate first but usually you would use the one that’s by itself already. like for example one of the equations was y = 8. you would already have your y solve for so you would just have to plug that in for y in the other equation. personally, i usually do x first unless one of the equations has either x or y by itself already. i think its easier to just do x first and then solve for y after that, but it just depends on what the equations are; sometimes it might be easier to just do y first. hope this helps!

5 0

3 years ago

Question in the attached below. @grujan50

denpristay [2]

The first answer is ''Linear pair postulate''

The second answer is ''Definition of suplementary''

The third answer is ''m∡ACB+45=180''

The fourth answer is ''Subtraction Property of Equality''

The fifth answer is ''∡ACB is obtuse''

The sixth answer is ''Definition of obtuse triangle''

Good luck!!!

7 0

4 years ago

A class of 49 students took a test. 17 students had an average score of 74. the other students had

Ann [662]

Answer: 83

Step-by-step explanation:

2816+1258/49

4074/49

582/7

6 0

2 years ago

Helllllllllpppppppp!!!!!

muminat

Answer:

X=10

A=84

Step-by-step explanation:

5x+34+2x+76=180

7x+110=180

7x=70

x=10

For A

5(10)+34

6 0

3 years ago