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Slope of AH = $\frac{a+2}{1}$ slope of BC = $-\frac{1}{p}\therefore p=a+2$ $\therefore C\left(\frac{18p-30}{p+1},\frac{15p-33}{p+1}\right)$  

slope of HC =  $\frac{16p-p^2-31}{16p-32}$  

slope of BC × slope of HC = -1 Þ p = 3 or 5

hence p = 3 is only possible value.

A salt bridge is a U-shaped tube that connects the two halves (oxidation and reduction) of the cell with an electrochemical cell, such as a galvanic cell. It allows the flow of ions in both halves and maintains the electricity to stay neutral.

Yes, IMT-CDL - Distance Learning admits students based on graduation scores. The institute admits students based on merit. Students can check the table below to know the detailed course eligibility:

${\displaystyle \underset{3}{\overset{x}{\int }}f\left(x\right)dx}={\left(\frac{f\left(x\right)}{x}\right)}^3\Rightarrow x^3{\displaystyle \underset{3}{\overset{x}{\int }}f\left(x\right)dx}=f^3\left(x\right),$ differentiating w.r.to x

$x^{3}f\left(x\right)+3x^2\frac{f^3\left(x\right)}{x^3}=3f^2\left(x\right)f\text{'}\left(x\right)\Rightarrow 3y^2\frac{dy}{dx}=x^{3}y=\frac{3y^3}{x}\Rightarrow 3xy\frac{dy}{dx}=x^4+3y^2$  

After solving we get  $y^2=\frac{x^4}{3}+c{x}^2$  also curve passes through (3, 3) Þ c = -2

$\therefore y^2=\frac{x^4}{3}-2x^2$ which passes through $\left(\alpha ,6\sqrt[]{10}\right)$ $\therefore \frac{{\alpha }^4-6{\alpha }^2}{3}=360\Rightarrow \alpha =6$  

According to question, we can write

$\omega =\pi =\sqrt[]{\frac{g}{\mathcal{l}}}\Rightarrow \mathcal{l}=\frac{g}{{\pi }^2}=\frac{9.8}{{\left(3.14\right)}^2}=0.99395=99.4cm$

In electrochemistry, metallic and electrolytic are two types of conductors.

The two branches of electrochemistry are: Electricity generated by chemical reactions and Chemical reactions that generate electricity.

${\displaystyle \underset{0}{\overset{1}{\int }}1.{\left(1-x^n\right)}^{2n+1}dx}$ using by parts we get

$\left(2n^2+n+1\right){\displaystyle \underset{0}{\overset{1}{\int }}{\left(1-x^n\right)}^{2n+1}dx=1177{\displaystyle \underset{0}{\overset{1}{\int }}{\left(1-x^n\right)}^{2n+1}dx}}$

$\Rightarrow 2n^2+n+1=1177\Rightarrow n=24or-\frac{49}{2}\therefore n=24$

According to question, we can write

$d=\sqrt[]{2hR}\Rightarrow \frac{d_2}{d_1}=\sqrt[]{\frac{h_2}{h_1}}\Rightarrow h_2={\left(\frac{d_2}{d_1}\right)}^2{h}_1={\left(\frac{2}{1}\right)}^2\times 125=500m$  

Increment in height of tower = h₂ – h₁ = 500 – 125 = 375 m

Coefficient of x in ${\left(1+x\right)}^p{\left(1-x\right)}^q={-}^p{C_0}^q{C}_1{+}^p{C_1}^q{C}_0=-3\Rightarrow p-q=3$  

              Coefficient of x² in ${\left(1+x\right)}^p{\left(1-x\right)}^q{=}^p{C_0}^q{C}_2{-}^p{C_1}^q{C}_1{-}^p{C_2}^q{C}_0=-5$  

$\Rightarrow \frac{q\left(q-1\right)}{2}-pq+\frac{p\left(p-1\right)}{2}=-5\Rightarrow \frac{q\left(q-1\right)}{2}-\left(q-3\right)q+\frac{\left(q-3\right)\left(q-4\right)}{2}=-5\Rightarrow q=11,p=8$  

              Coefficient of x³ in ${\left(1+x\right)}^8{\left(1-x\right)}^{11}={-}^{11}{C}_3{+}^8{C_1}^{11}{C}_2{-}^8{C_2}^{11}{C}_1{+}^8{C}_3=23$  

$x^8-x^7-x^6+x^5+3x^4-4x^3-2x^2+4x-1=0$  

$\Rightarrow x^7\left(x-1\right)-x^5\left(x-1\right)+3x^3\left(x-1\right)-x\left(x^2-1\right)+2x\left(1-x\right)+\left(x-1\right)=0$  

$\Rightarrow \left(x-1\right)\left(x^2-1\right)\left(x^5+3x-1\right)=0\therefore x=\pm 1$  are roots of above equation and x⁵ + 3x – 1 is a monotonic term hence vanishes at exactly one value of x other then 1 or -1.

 3 real roots.

Kindly go through the solution 

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Given series $\left\{3\times 1\right\},\left\{3\times 2,3\times 3,3\times 4\right\},\left\{3\times 5,3\times 6,3\times 7,3\times 8,3\times 9\right\}.........$  

$\therefore$  11^th set will have 1 + (10)2 = 21 terms

Also up to 10^th set total 3 × k type terms will be 1 + 3 + 5 + ……… +19 = 100 terms

$\therefore Set11=\left\{3\times 101,3\times 102,......3\times 121\right\}$ $\therefore$ Sum of elements = 3 × (101 + 102 + ….+121)

$=\frac{3\times 222\times 21}{2}=6993.$   

Factors of 36 = 2².3².1

Five-digit combinations can be 

(1, 2, 3, 3), (1, 4, 3, 1), (1, 9, 2, 1), (1, 4, 9, 11), (1, 2, 3, 6, 1), (1, 6, 1, 1)

i.e., total numbers  $\frac{5!5!}{2!2!}+\frac{5!}{2!2!}+\frac{5!}{2!2!}+\frac{5!}{3!}+\frac{5!}{2!}+\frac{5!}{3!2!}=\left(30\times 3\right)+20+60+10=180.$  

If currents are flowing in same direction, magnetic field will cancel each other, so the currents must flowing in opposite direction

$B_P=\frac{{\mu }_{0}I}{2\pi r}\times 2$

$\Rightarrow 300\times 10^{-6}=\frac{4\pi \times 10^{-7}}{2\pi \times 4\times 10^{-2}}\times 2$                                                                           

$\Rightarrow$ I = 30 A

Let a and b are the roots of $\left(p^2+q^2\right)x^2-2q\left(p+r\right)x+q^2+r^2=0$  

  $\therefore \alpha +\beta >0and\alpha \beta >0$ Also, it has a common root with x² + 2x – 8 = 0

$\therefore$  The common root between above two equations is 4.

  $\Rightarrow 16\left(p^2+q^2\right)-8q\left(p+r\right)+q^2+r^2=0\Rightarrow \left(16p^2-8pq+q^2\right)+\left(16q^2-8qr+r^2\right)=0$  

      $\Rightarrow {\left(4p-q\right)}^2+{\left(4q-r\right)}^2=0\Rightarrow q=4pandr=16p\therefore \frac{q^2+r^2}{p^2}=\frac{16p^2+256p^2}{p^2}=272$  

Candidates looking for admission at IMT-CDL - Distance Learning must complete at least a graduation degree. The basic eligibility for PGDM courses is to complete graduation. For Certificate courses, candidates must have at least three years of experience. Apart from the work experience, students who belong to the final semester of PGDM/PGDM-Executive/ PGCM at IMT CDL can apply for certificate courses, excluding the Proficiency Certificate Programme in Financial Management course.

Yes, Mathematics is compulsory to get admission in the MCA programme provided at Hillwoods Academy of Higher Education. Students need to have studied the subject in either Class 12 or during their Bachelor's degree. Eligible students can apply online for Hillwoods Academy of Higher Education MCA admission.

The institute offers three PGDM and five Certificate courses. Apart from the PGDM and Certificate courses, the institute also offers an Online MDP course. The basic eligibility criteria for admission are to complete graduation. For Certificate courses, candidates must have at least three years of experience. Apart from the work experience, students who belong to the final semester of PGDM/PGDM-Executive/ PGCM at IMT CDL can apply for certificate courses, excluding the Proficiency Certificate Programme in Financial Management course.