Recent content by skeeter

using your sub \( u = \sqrt{y} \) ... \( \displaystyle 2 \int \dfrac{1}{\cos^2{\sqrt{y}}} \cdot \dfrac{dy}{2\sqrt{y}} = 2\int \sec^2{u} \, du \)

vertical distance between two points = $|\Delta y|$

prime factorization … $35280 = 2^4 \cdot 3^2 \cdot 5 \cdot 7^2$ $4158 = 2 \cdot 3^3 \cdot 7 \cdot 11$ greatest common divisor includes the least power of all common factors … $2 \cdot 3^2 \cdot 7 = 126$ least common multiple includes the greatest power of all common factors and the factors...

$u = \sqrt{x} \implies du = \dfrac{dx}{2\sqrt{x}}$ $\displaystyle 3 \int \dfrac{dx}{x(2-\sqrt{x})} = 2 \cdot 3\int \dfrac{dx}{2\sqrt{x}(2\sqrt{x} - x)}$ substitute ... $\displaystyle 6\int \dfrac{du}{2u - u^2} = -6 \int \dfrac{du}{(u^2 - 2u + 1) - 1} = -6 \int \dfrac{du}{(u-1)^2 - 1}$ since...

$\displaystyle 3\int \dfrac{dx}{(x+2)\sqrt{x^2+4x+4-1}} = 3\int \dfrac{dx}{(x+2)\sqrt{(x+2)^2-1}}$ let $u = x+2 \implies du = dx$ ... $\displaystyle 3\int \dfrac{1}{u\sqrt{u^2-1}} \, du$ substitution again ... $v = \sqrt{u^2-1}$ continue ...

https://www.physicsforums.com/attachments/312509._xfImport https://mathhelpboards.com/help/forum_rules/

$M$ = 8kg, $m$ = 5kg, $T$ is the tension force in the string $Mg - T = Ma$ $T - mg = ma$ Solve the system of equations for $a$, the magnitude of the acceleration for both masses. Once you find that acceleration, you can find the upward velocity of the smaller mass when the larger one hits the...

$\dfrac{5280 \, ft}{mile} \cdot \dfrac{12 \, in}{ft} \cdot \dfrac{2.54 \, cm}{in} \cdot \dfrac{1 \, m}{100 \, cm} \cdot \dfrac{1 \, km}{1000 \, m}$

Purchasing 600,000 of equity from company C is still 1.2% of company A since C is contained within A. Still not seeing how the equity shares of A held by B and C make any difference.

1.2% of Company A’s value of 50 million is 600,000 not clear what Company C’s equity of 30 million has to do with that 1.2%

Maybe you meant ... $\displaystyle \int 3x+5 \, dx = \dfrac{3}{2}x^2 + 5x + C$ ? in any case, watch the video

no ... you found the derivative, not the antiderivative. $\displaystyle \int \dfrac{3}{2}x^2 + 5x + C \, dx = \dfrac{x^3}{2} + \dfrac{5}{2}x^2 + Cx + K$

$\dfrac{1}{2}gt^2 - v_{y_0} \cdot t + \Delta y = 0 \implies t = \dfrac{v_{y_0} + \sqrt{(v_{y_0})^2 - 2g\Delta y}}{g}$ for the given values, $t \approx 3.7 \, sec$ $\Delta x = v_{x_0} \cdot t \approx 55.5 \, m$ due South of the drop position.

$\Delta x = v_{x_0} \cdot t$ $\Delta y = v_{y_0} \cdot t - \dfrac{1}{2}gt^2$ solve the quadratic for $t$, then calculate $\Delta x$