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No, direct admissions are not possible in Maya Academy of Advanced Creativity. Maya Academy of Advanced Cinematics, Dehradun application process 2025 is conducted online. A step-by-step process for filling out the online application form is mentioned below:

• Step 1: Click on the link provided here.
• Step 2: Scroll down and find the registration form.
• Step 3: Fill up the registration form and submit it.

There are two types of categories : sponsored and non-sponsored (based on merit). Although the precise distribution isn't made public, the non-sponsored category usually receives a bigger share. Also, there is some merit-based flexibility, sponsored category admissions come with a higher tuition cost.

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l}Givenplanesare:\\ \overrightarrow{r}.\left(\widehat{i}-3\widehat{j}\right)-6=0\Rightarrow x+3y-6=0\dots \left(i\right)\\ and\overrightarrow{r}.\left(3\widehat{i}-\widehat{j}-4\widehat{k}\right)=0\Rightarrow 3x-y-4z=0\dots \left(ii\right)\\ Equationoftheplanepassingthroughthelineofintersectionofplane\left(i\right)and\left(ii\right)is\\ \left(x+3y-6\right)+k\left(3x-y-4z\right)=0\dots \left(iii\right)\\ \left(1+3k\right)x+\left(3-k\right)y-4kz-6=0\\ Perpendiculardistancefromorigin\\ \Rightarrow \left|\frac{-6}{\sqrt[]{{\left(1+3k\right)}^2+{\left(3-k\right)}^2+{\left(-4k\right)}^2}}\right|=1\\ \Rightarrow \frac{36}{1+9k^2+6k+9+k^2-6k+16k^2}=1\left[Squaringbothsides\right]\\ \Rightarrow \frac{36}{26k^2+10}=1\Rightarrow 26k^2+10=36\\ \Rightarrow 26k^2=26\Rightarrow k^2=1\therefore k=\pm 1\\ Puttingthevalueofkineq.\left(iii\right)weget,\\ \left(x+3y-6\right)\pm \left(3x-y-4z\right)=0\\ \Rightarrow x+3y-6+3x-y-4z=0andx+3y-6-3x+y+4z=0\\ \Rightarrow 4x+2y-4z-6=0and-2x+4y+4z-6=0\\ Hence,therequiredequationare\\ 4x+2y-4z-6=0and-2x+4y+4z-6=0\end{array}$

Maya Academy of Advanced Cinematics, Dehradun application process 2025 is conducted online. A step-by-step process for filling out the online application form is mentioned below:

• Step 1: Click on the link provided here.
• Step 2: Scroll down and find the registration form.
• Step 3: Fill up the registration form and submit it.

No, Maya Academy of Advanced Cinematics, Dehradun admissions are not hard especially for hardworking students.

A degree from ICRI Bangalore - Institute of Clinical Research India Bangalore is highly valued. Reputable organizations frequently participate in the placements process to offer candidates for job openings once they have completed their undergraduate or postgraduate course. 

Candidates may choose to pursue interesting jobs after graduation from ICRI Bangalore - Institute of Clinical Research India Bangalore. The strong network of former students shows that the university produces future capable leaders. 

Maya Academy of Advanced Cinematics, Dehradun admission (registration form) is currently open. The institute offers Certificate courses. To get admitted to MAAC, candidates must meet the eligibility criteria set by the institute. The application process at Maya Academy of Advanced Cinematics, Dehradun is conducted online. The documents that shall be kept ready for the application process are:

• Scanned Class 10 and Class 12 mark sheet
• Leaving Certificate
• Aadhaar Card
• Passport-size photograph
• Signature

No, the fees vary from course to course. The courses offered have different categories and specialisations hence the fees differs. For e.g., fees for BJMC Evening is lesser than regular BJMC. All the courses have different fee brackets for sponsored and non-sponsored category.

This is a True or False type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}If\overrightarrow{a}.\overrightarrow{b}=0then\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|cos90^0\\ Sotheanglebetweentheadjacentsidesoftherhombusshouldbe90^{0}whichisnotpossible.\\ HencethegivenstatementisFalse.\end{array}$

Yes, it is mandatory to give Personal Interview to take admission in Amity School of Communication. It is generally conducted by the faculty and is to assess the candidate's knowledge and communication skills. However, candidates who are granted direct admission because of their educational qualifications may not give the PI.

The admission process in Amity School of Communication includes three main stages. The first one is the registration process that have to be completed by the candidates and information should be clear added to avoid rejection. After this, the applications are reviewed and applicants have a Personal Interview round which passing which is necessary after which they get an admission call and have to pay the fee to secure their seat. For MBA there is an additional written test before PI.

While comparing Silver Oak University MBA programme with other institutes, students should consider key factors such as total tuition fees, entrance exam requirements, specialisations offered, faculty quality, placement records, infrastructure, and student reviews. These elements help evaluate the overall value and suitability of the programme for individual career goals.

The overall vibe of Maya Academy of Advanced Cinematics is quite good. Maya Academy of Advanced Cinematics offers several facilities for its students. The facilities include medical and sports facilities for the students. The institute has a sports committee. A senior faculty member is nominated as the convener of the committee along with a selected body comprising faculty members and students. The committee conducts all types of sports activities.

The fee for the courses offered by ICRI Bangalore are as follows: 

Candidates who clear the admission process are eligible to pay the course fee and secure their seat. 

Indeed, the Amity School of Communication (ASCO) recognizes the Common Admission Test score for entry into its Master of Business Administration course. Candidates who achieve a percentile of 75 or higher in the Common Admission Test are eligible to participate in the selection process at Amity School of Communication. In addition to the Common Admission Test, Amity School of Communication also considers other entrance exams such as GMAT, NMAT, and MAT.

Amity School of Communication is a renowned media institute in and also offers good placements to students. For UG and PG courses the average return or placement package is INR 3 LPA and INR 4 LPA respectively. Some big media companies such as Zee, Network 18, AdFactors PR also come to ASCO for hiring.

This is a True or False type Questions as classified in NCERT Exemplar

Sol:

       $\begin{array}{l}{\left(\overrightarrow{a}+\overrightarrow{b}\right)}^2=\left(\overrightarrow{a}+\overrightarrow{b}\right).\left(\overrightarrow{a}+\overrightarrow{b}\right)\\ ={\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+2\overrightarrow{a}.\overrightarrow{b}\\ HencethegivenstatementisFalse.\end{array}$

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l}Thegivenplanesare\\ ax+by=0\dots \left(i\right)\\ z=0\dots \left(ii\right)\\ Equationofanyplanepassingthroughthelineofintersectionofplane\left(i\right)and\left(ii\right)is\\ \left(ax+by\right)+kz=0\Rightarrow ax+by+kz=0\dots \left(iii\right)\\ Dividingbothsidesby\sqrt[]{a^2+b^2+k^2},weget\\ \frac{a}{\sqrt[]{a^2+b^2+k^2}}x+\frac{b}{\sqrt[]{a^2+b^2+k^2}}y+\frac{k}{\sqrt[]{a^2+b^2+k^2}}z=0\\ \therefore Directioncosinesofthenormaltotheplaneare\\ \frac{a}{\sqrt[]{a^2+b^2+k^2}},\frac{b}{\sqrt[]{a^2+b^2+k^2}},\frac{k}{\sqrt[]{a^2+b^2+k^2}}\\ andthedirectioncosinesoftheplane\left(i\right)are\\ \frac{a}{\sqrt[]{a^2+b^2}},\frac{b}{\sqrt[]{a^2+b^2}},0\\ Since,\alpha istheanglebetweentheplanes\left(i\right)and\left(iii\right),weget\\ cos\alpha =\frac{a.a+b.b+k.0}{\sqrt[]{a^2+b^2+k^2}.\sqrt[]{a^2+b^2}}\\ \Rightarrow cos\alpha =\frac{a^2+b^2}{\sqrt[]{a^2+b^2+k^2}.\sqrt[]{a^2+b^2}}\\ \Rightarrow cos\alpha =\frac{\sqrt[]{a^2+b^2}}{\sqrt[]{a^2+b^2+k^2}}\Rightarrow co{s}^2\alpha =\frac{a^2+b^2}{a^2+b^2+k^2}\\ \Rightarrow \left(a^2+b^2+k^2\right)co{s}^2\alpha =a^2+b^2\\ \Rightarrow a^{2}co{s}^2\alpha +b^{2}co{s}^2\alpha +k^{2}co{s}^2\alpha =a^2+b^2\\ \Rightarrow k^{2}co{s}^2\alpha =a^2-a^{2}co{s}^2\alpha +b^2-b^{2}co{s}^2\alpha \\ \Rightarrow k^{2}co{s}^2\alpha =a^2\left(1-co{s}^2\alpha \right)+b^2\left(1-co{s}^2\alpha \right)\\ \Rightarrow k^{2}co{s}^2\alpha =a^{2}si{n}^2\alpha +b^{2}si{n}^2\alpha \\ \Rightarrow k^{2}co{s}^2\alpha =\left(a^2+b^2\right)si{n}^2\alpha \\ \Rightarrow k^2=\left(a^2+b^2\right)\frac{si{n}^2\alpha }{co{s}^2\alpha }\Rightarrow k=\pm \sqrt[]{a^2+b^2}.tan\alpha \\ Puttingthevalueofkineq.\left(iii\right)weget\\ ax+by\pm \left(\sqrt[]{a^2+b^2}.tan\alpha \right)z=0whichistherequiredEquationofplane.\\ Henceproved.\end{array}$