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14 Views

how can i give jee mains examination while i am not comfortable in computer........is it essential for me to join any computer class for being comfortable

Tanya Gupta 15th Feb, 2024

Hello,

If you are not comfortable in using computers for the JEE Mains exam, please consider these tips:

Practice online mock tests to get used to the format.

Learn basic computer skills through tutorials.

Use tutorials provided by the exam authority.

Seek help from friends.

Hope this helps you,

Thank

18 Views

when will they conduct jee mains examination for 2024 first year students

Tanya Gupta 28th Feb, 2024

Hello,

Session 1 is scheduled from January 24th, 2024, to February 1st, 2024. Session 2 exams are set to commence on April 1st, 2024, and conclude on April 15th. The agency will soon release the notification for the exams. But one can expect examination between this period that is mentioned

80 Views

I have passed all my cbse boards examination but not happy with result if I will appear in improvement exam then will I be eligible for afmc

Shivanshu 27th Jan, 2024

Dear aspirant !!

Hope you are doing well ! See , it doesn't matter whether you give improvement exam or not . Your result must be pass . You will be Eligible don't worry about that ,just focus on your preparation.

The criteria is ;-

Candidates must have passed 10+2

8 Views

Question : If $\left(x+\frac{1}{x}\right)=5$, and $x>1$, what is the value of $\left(x^8-\frac{1}{x^8}\right)?$

Option 1: $60605 \sqrt{21}$

Option 2: $60615 \sqrt{21}$

Option 3: $60705 \sqrt{21}$

Option 4: $60725 \sqrt{21}$

Team Careers360 24th Jan, 2024

Correct Answer: $60605 \sqrt{21}$


Solution : Given: $\left(x+\frac{1}{x}\right)=5$--------------(i)
Now, $\left(x^8-\frac{1}{x^8}\right)$
$=(x^4-\frac{1}{x^4})$$(x^4+\frac{1}{x^4})$
$=(x^2-\frac{1}{x^2})$$(x^2+\frac{1}{x^2})$$(x^4+\frac{1}{x^4})$
$= (x-\frac{1}{x})$$(x+\frac{1}{x})$$(x^2+\frac{1}{x^2})$$(x^4+\frac{1}{x^4})$----------(ii)
Squaring equation (i), we get,
$(x+\frac{1}{x})^2 = 5^2$
$⇒(x^2+\frac{1}{x^2}+2) = 25$
$⇒(x^2+\frac{1}{x^2}) = 23$-----------------(iii)
Squaring equation (iii), we get,
$(x^2+\frac{1}{x^2})^2 = 23^2$
$⇒(x^4+\frac{1}{x^4}+2) = 529$
$⇒(x^4+\frac{1}{x^4}) = 527$---------------(iv)
Subtracting 2 on both sides of equation (iii),

249 Views

is there a separate examination for JIPMER for the UG MBBS EXAM

mjaisinghani62 22nd Jan, 2024

Hello aspirant .

The entrance exam for ug mbbs is Neet ug.

Previously JIPMER entrance exam was there for admission  to mbbs or medical  courses . It is now replaced with neet ug.

Jawaharlal Institute of post graduate medical education and research was established in 1956.

Jipmer is ranked number

210 Views

Whats the fee structure for msc clinical embryology and assisted reproductive technology in Reva university? Also is there any entrance examination for admission to this course??

mjaisinghani62 25th Jan, 2024

Hello Annex


For msc in clinical embryology from reva university Fees charges are 80,000 for two years.

Msc in clinical embryology and assisted reproductive technology deals with study of in vitro fertilization .

( IVF)

These days invitro fertilization has proved a boon to innumerable infertile couples . Over the

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