Sum of divisors of 2 5 ·3 7 ·5 3 · 7 2 is –

The condition for the lines $c subscript 1 over r equals a subscript 1 c o s space theta plus b subscript 1 s i n space theta$ and $c subscript 2 over r equals a subscript 2 c o s space theta plus b subscript 2 s i n space theta$ to be perpendicular is

If f : R →R; f(x) = sin x + x, then the value of $not stretchy integral subscript 0 end subscript superscript pi end superscript blank$ (f^-1 (x)) dx, is equal to

The polar equation of the straight line with intercepts 'a' and 'b' on the rays $theta equals 0$ and $theta equals fraction numerator pi over denominator 2 end fraction$ respectively is

The polar equation of the straight line parallel to the initial line and at a distance of 4 units above the initial line is

The polar equation of $x to the power of 3 end exponent plus y to the power of 3 end exponent equals 3$ axy is

If x, y, z are integers and x $greater or equal than$ 0, y $greater or equal than$ 1, z $greater or equal than$ 2, x + y + z = 15, then the number of values of the ordered triplet (x, y, z) is -

If $alpha not equal to beta comma alpha 2 equals 5 alpha minus 3 comma beta 2 times equals 5 beta minus 3 comma$ then the equation whose roots are $alpha divided by beta straight & beta divided by alpha$

Let p, q $element of$ {1, 2, 3, 4}. Then number of equation of the form px² + qx + 1 = 0, having real roots, is

Here we used the concept of quadratic equations and solved the problem. We also understood the concept of discriminant and used it in the solution to find the intervals. Therefore, total 7 seven solutions are possible so $7$ equations can be formed.

Let p, q $element of$ {1, 2, 3, 4}. Then number of equation of the form px² + qx + 1 = 0, having real roots, is

ax² + bx + c = 0 has real and distinct roots null. Further a > 0, b < 0 and c < 0, then –

Here we used the concept of quadratic equations and solved the problem. The concept of sum of roots and product of roots was also used here. Therefore, 0 < ∣α∣ < β.

ax² + bx + c = 0 has real and distinct roots null. Further a > 0, b < 0 and c < 0, then –

Here we used the concept of quadratic equations and solved the problem. The concept of sum of roots and product of roots was also used here. Therefore, 0 < ∣α∣ < β.

The cartesian equation of $r squared c o s space 2 theta equals a squared$ is

Here we used the concept of quadratic equations and solved the problem. We also understood the concept of trigonometric ratios and used the formula to find the equation. So the equation is $x to the power of 2 end exponent minus y to the power of 2 end exponent equals a to the power of 2 end exponent$ .

The cartesian equation of $r squared c o s space 2 theta equals a squared$ is

The castesian equation of $r c o s invisible function application left parenthesis theta minus alpha right parenthesis equals p$ is

Here we used the concept of quadratic equations and solved the problem. We also understood the concept of trigonometric ratios and used the formula to find the equation. So the equation is $x c o s space alpha plus y s i n space alpha equals p$ .

The castesian equation of $r c o s invisible function application left parenthesis theta minus alpha right parenthesis equals p$ is

Here we used the concept of quadratic equations and solved the problem. We also understood the concept of trigonometric ratios and used the formula to find the equation. So the equation is $x c o s space alpha plus y s i n space alpha equals p$ .

The equation of the directrix of the conic whose length of the latusrectum is 5 and eccenticity is 1/2 is

The equation of the directrix of the conic $r C o s to the power of 2 end exponent invisible function application open parentheses fraction numerator theta over denominator 2 end fraction close parentheses equals 5$ is

The equation of the circle touching the initial line at pole and radius 2 is