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Trouble here in the below partial fraction (Bug) $\frac{5x^2+1}{(3x+2)(x^2+3)}$ One factor in the denominator is a quadratic expression Split this into two parts A&B $\frac{5x^2+1}{(3x+2)(x^2+3)}=\frac{A}{(3x+2)}+\frac{Bx+c}{(x^2+3)}$...

Thanks Mark, nicely explained

Hello Everyone , I need some help in solving this partial fraction $\frac{x^2}{(x-2)(x+3)(x-1)}$ I am using this method in which the partial fraction is broken into 3 parts namely A,B &C $\frac{x^2}{(x-2)(x+3)(x-1)}=\frac{A}{(x-2)}+\frac{B}{(x+3)}+\frac{C}{(x-1)}$...

An explanation on this please

That's what was exactly missing here, ADFC is the cyclic quadrilateral and from there AFD= FDC and ADF= DFC ADF + FDC=ADC AFD+DFC=AFC AFD+DFC+ADF+FDC=180 2ADF + 2FDC = 180 dividing both sides by 2 ADF+FDC=90 ADC=90 Taking it the alternative way, Both squares and rectangles have...

:D A little bit confused here, A cyclic quadrilateral is a quadrilateral whose vertices all touch the circumference of a circle. But here there is no circle around the figure Many Thanks :)

180 degrees , But a cyclic quadrilateral is not to be seen here (Thinking)

The triangle ADC exactly one half of the area of the triangle ABC and the parallelogram ADCF is twice the area of triangle ADC and triangle ABC is twice the area of triangle ABC Now what remains is, Show that if $DE=AE$, then $\angle ADC=90^{\circ}$ Many THanks :)

Workings $\triangle ADE \cong \triangle CFE \left(AAS\right)$ $\angle AED = \angle CEF $( vertically opposite angles ) $\angle CFE= \angle EDA $( alternate angles ) $AE=EC $( E midpoint ) $ii.$ADCF is a parallelogram because diagonals bisect each other. Where is help needed How should...

Re: ACT problem To find the x intercept of the graph of the function $y = x^2 – 4x + 4$ We may either use the completing the square method or we may use the formula $x=\frac{-b}{2a}$ The easiest way is to use $x=\frac{-b}{2a}$ From $y = x^2 – 4x + 4$ which is in the form of $ax^2+bx+c$ we...

The first term is 3 & the common difference as calculated above is +8. And I guess you are supposed to find the sum of first 20 terms of the above arithmetic progression in which all the 20 terms are odd numbers (Thinking)

From the first 20 term the number of odd numbers would be 10. (Thinking)

The third term $T_3=3+(3-1)4=8+3=11$ The difference between the first and the third term is $3 {\underbrace{\phantom{2d) + (3e}}_{\text{+8}}} 11$ The fifth term $T_5=3+(5-1)4=16+3=19$ The difference between the first and the third term is $11 {\underbrace{\phantom{2d) + (3e}}_{\text{+8}}}...

First term of the progression is 3 & the common difference is 4 Find the sum of the first 20 terms of the progression that is obtained by removing the terms in the even positions of the given progressions, such as the second term,fourh term, sixth term. Formula preferences For the sum of an...

I was just taking a look again at this problem and how can we say that $\measuredangle XEB$ = 90 degrees , I know that $\measuredangle XEB$ = 90 degrees but how can we prove that ? as nowhere in the problem mentions that line XE is a perpendicular drawn $\measuredangle ADC$ can be proved 90...