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
$\frac{5x^2+1}{(3x+2)(x^2+3)}=\frac{A}{(3x+2)}+\frac{Bx+c}{(x^2+3)}$...
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
$\frac{x^2}{(x-2)(x+3)(x-1)}=\frac{A}{(x-2)}+\frac{B}{(x+3)}+\frac{C}{(x-1)}$...
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 :)
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)
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...