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MIT School of Distance Education offers courses at PG, Diploma and certificate levels. The school offers two-year PGDM/ PG Diploma and 15-month Executive PGDM programmes. Besides, MITSDE also offers one-year certificate and six-months professional certificate courses in Management. Interested candidates can fill out the application form for preferred course online.

The SI unit of the electrical power is Watt which is symbolized as W. One watt is equal to one joule per second (J/s). In the context of current and voltage, one watt is the power consumed when one ampere of the current flows through the potential difference of one volt.

The Bihar BEd CET (Common Entrance Test) exam is conducted in 11 cities across Bihar. Here's a list of the cities where the exam is typically held: Patna, Muzaffarpur, Gaya, Darbhanga, Bhagalpur, Purnea, Ara, Chapra, Munger, Begusarai, and Rosera. Lalit Narayan Mithila University (LNMU) is the conducting body for this exam

I don't think so. The basic qualification requirements for both girls and boys are the same. Though age, height & weight, or physical fitness & health parameters for both genders can differ. Based on the college or course you are choosing, you can check the accurate eligibility requirements from the institute's official website.

Hope this helps!!!

The programme fee for Post Graduate Certificate in Management or PGCM is INR 64,000. This information is sourced from official website/ sanctioning body and is subject to change. Aspirants have to pay the programme fee at the time of admission to confirm their seat. Candidates can deposit the fee via any of the online payment modes.

$y^{1/4}+\frac{1}{y^{1/4}}=2x$

${\left(y^{1/4}\right)}^2-2x{y}^{1/4}+1=0$    

=> ${\left(y^{1/4}\right)}^2-2x{y}^{1/4}+1=0$

$\Rightarrow \frac{dy}{dx}=\frac{4y}{\sqrt[]{x^2-1}}..........\left(i\right)$

$\Rightarrow \left(x^2-1\right)\frac{d^{2}y}{d{x}^2}+\frac{xdy}{dx}-16y=0from\left(i\right)$      

=>α = 1, β = -16

|α - β| = 17

${\left(3x^2+4x+3\right)}^2-\left(k+1\right)\left(3x^2+4x+3\right)\left(3x^2+4x+2\right)+k{\left(3x^2+4x+2\right)}^2=0$

 Let $3x^2+4x+3=a\&3x^2+4x+2=b\Rightarrow a-1$  

So, (i) becomes a² – (k + 1)ab + kb² = 0

(a – kb) (a – b) = 0 Þ a = kb or a = b ® not possible

->3x² + 4x + 3 = k (3x² + 4x + 2)

For real roots $D\ge 0$  

$16{\left(k-1\right)}^2-12\left(k-1\right)\left(2k-3\right)\ge 0$     

So, $k\in \left(1,\frac{5}{2}\right]$  

The admission process for Primus B School involves multiple steps, starting from application, followed by selection rounds, and admission confirmation. Prospective students must satisfy the admission criteria of the desired course. Those who meet the eligibility criteria can then apply online by submitting the application form. The submitted application is then reviewed by the admission team of the Primus B School.  Selected candidates are then required to pay the admission fees of the course.

Aspirants are shortlisted for certificate courses based on their merit. MIT School of Distance Education considers qualifying exam scores of the candidates for admission. The minimum qualification required for PG certificate courses is graduation, whereas Class 10/ Class 12 scores for professional certificate programmes. Once selected, aspirants have to confirm their seat by paying the course fee.

Let $l={\displaystyle \int \frac{dx}{{\left(x^2+x+1\right)}^2}.............\left(i\right)}$

Now ${\displaystyle \int \frac{dx}{x^2+x+1}={\displaystyle \int \left(\frac{1}{x^2+x+1}\right)dx}}$

by integration by parts

=> ${\displaystyle \int \frac{dx}{{\left(x^2+x+1\right)}^2}=\frac{4}{3\sqrt[]{2}}ta{n}^{-1}\frac{2x+1}{\sqrt[]{3}}+\frac{1}{3}\frac{\left(2x+1\right)}{\left(x^2+x+1\right)}+c}$

$\Rightarrow a=\frac{4}{3\sqrt[]{3}},b=\frac{1}{3}$

Now, $9\left(\sqrt[]{3}a+b\right)=9\left(\frac{4}{3}+\frac{1}{3}\right)=15$

There are five important sections included in the Bihar BEd CET Syllabus: General English Comprehension, General Hindi, Teaching-Learning Environment in Schools, General Awareness, and Logical & Analytical Reasoning. General Awareness section includes 40 questions for 40 marks in the Bihar BEd CET exam making it the mosts scoring section in the examination.

Ge (Z = 32)

^{$=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^2$}         

$m_{\mathcal{l}}=0$       (for 4s, 3s, 2s, 1s) 4 orbital

$m_{\mathcal{l}}$       = 0 (one p orbital of 2p and 3p)

$m_{\mathcal{l}}$ = 0 (one d orbital)

Total orbitals = 7

Ans. = 7

Equation of the plane will be $\left\{\overrightarrow{r}.\left(\widehat{i}+\widehat{j}+\widehat{k}\right)-1\right\}+\lambda \left\{\overrightarrow{r}.\left(2\widehat{i}+3\widehat{j}-\widehat{k}\right)+4\right\}=0$   

$\overrightarrow{r}.\left\{\left(1+2\lambda \right)i+\left(1+3\lambda \right)\widehat{j}+\left(1-\lambda \right)\widehat{k}\right\}+\left(4\lambda -1\right)=0$               

-> $\left(1+2\lambda \right)x+\left(1+3\lambda \right)y+\left(1-\lambda \right)z+\left(4\lambda -1\right)=0.........\left(i\right)$

(i) is parallel to x-axis so its d.r.s will be (1, 0, 0)

-> $1+2\lambda =0so\lambda =\frac{-1}{2}$

Hence required equation will be

$\overrightarrow{r}\left\{-\frac{1}{2}\widehat{j}.+\frac{3}{2}\widehat{k}\right\}+\left(-3\right)=0$

$\overrightarrow{r}.\left(\widehat{j}-3\widehat{k}\right)+6=0$              

The total tuition fee for BBA at Academy of Aviation & Engineering, Bangalore costs around 3,00,000 LPA to INR 2,50,000 LPA. Tuition fee is paid in installements, at the beginning of each academic cycle. Other than this, there are other charges such as hostel and mess fees. While admission fee is a one-time payment that is paid when you take admission into the programme. For the ease of understanding, students can split the fee expenses into: one time payment and installment type payments.

Variance = $\frac{\sum {\left(x_i-\overline{x}\right)}^2}{n}=\frac{\sum \left({x_i}^2+{\overline{x}}^2-2\overline{x}xi\right)}{n}$  

$=\frac{\sum {x_i}^2+{\overline{x}}^2\sum 1-2\overline{x}\sum x_i}{n}$

$=\frac{\frac{n\left(n+1\right)\left(2n+1\right)}{6}+{\left(\frac{n\left(n+1\right)}{2n}\right)}^2.n-\frac{2\left(n+1\right)}{2n}.\frac{n\left(n+1\right)}{2}}{n}$ $=\frac{n^2-1}{12}$

Now, $\frac{n^2-1}{12}=14son=13$

MIT School of Distance Education offers six-month professional certificate programmes. Candidates who have completed Class 10/ Class 12 are eligible to apply for these courses. The school offers professional certificate in Project Management, Human Resource,  Marketing,  Operations, etc.

The Bihar BEd CET syllabus covers five important areas: General English/Sanskrit Comprehension, General Hindi, Logical and Analytical Reasoning, General Awareness, and Teaching-Learning Environment in Schools. These areas include topics like synonyms and antonyms, one-word substitution, geography, and current affairs. The exam consists of 120 multiple-choice questions, with each section containing objective-type questions.