Cymbal:
Determine the type of conic section represented by the equation x²/k + y²/(k-9) = 1 in each of the following cases.
(i.) k > 9 ( ii.) 0 < k > 9 and ( iii. ) k < 0. For the case k = 25 ,find the focus or foci and the equation of the directrix or directrixes of the Conic.
Tag: laplacian,agentofchange1,jackpot,benji93,arithmetic,detj4u.

tonywirelex:
maths105,100level Abu zaria test.
Question;
given that J=1
sint=m
α=β/root of J
solve
∫α't/(Jpower2 - mpower2)0.452/0.904

thankyouJesus:
your question is not clear, make it understandable.

tonywirelex:
maths105,100level Abu zaria test.
Question;
given that J=1
sint=m
α=β/root of J
solve
∫α't/(Jpower2 - mpower2)0.452/0.904

thankyouJesus:
root of j, which is 1 is equal to 1.
Alpha = beta, I am lost, I can't read your question.

tonywirelex:
abu test

Cymbal:
Determine the type of conic section represented by the equation x²/k + y²/(k-9) = 1 in each of the following cases.
(i.) k > 9 ( ii.) 0 < k > 9 and ( iii. ) k < 0. For the case k = 25 ,find the focus or foci and the equation of the directrix or directrixes of the Conic.
Tag: laplacian,agentofchange1,jackpot,benji93,arithmetic,detj4u.

tonywirelex:
maths105,100level Abu zaria test.
Question;
given that J=1
sint=m
α=β/root of J
solve
∫α't/(Jpower2 - mpower2)0.452/0.904

Kendzyma:
Lol,exam question num 4.. u skull in oau rit?

Arithmetic:

(i) Ellipse.
(ii) Hyperbola.
(iii) does not exist.
x²/25 + y²/16 = 1; comparing with: x²/a²+y²/b²=1,
a=5, b=4. To find the eccentricity, we use:
b²=a²(1-e²), e=3/5.
Focus, S(±ae, 0);
S(3,0) and S¹(-3,0).
To obtain directrices (±a/e,0), which are:
(25/3) and (-25/3)

Kendzyma:
tho the question don't specify what we are integrating with respect to..BT let me guess...
..j=1
sint=m,t=arcsinm
®=alpha,©=beta
...int ®t/√(j^2-m^2).dm
int ©arcsinm/√(j^2-m^2) .dm
.
©/1 int arcsinm/√1-m^2dm..........(since j=1)
let u = arcsinm
dm=√1-m^2du...
int udu
u^2/2+c
©sin^2m/2+c.........dats if alpha' doesn't mean diferential....

Arithmetic:

(i) Ellipse.
(ii) Hyperbola.
(iii) does not exist.
x²/25 + y²/16 = 1; comparing with: x²/a²+y²/b²=1,
a=5, b=4. To find the eccentricity, we use:
b²=a²(1-e²), e=3/5.
Focus, S(±ae, 0);
S(3,0) and S¹(-3,0).
To obtain directrices (±a/e,0), which are:
(25/3) and (-25/3)

Arithmetic:
Since P is equidistant from A(a,0) and B(-a,0);
=> PA² = (x-a)² + y²
=> PB² = (x+a)² +y²
PA²-PB² = 2k².
=> (x-a)² + y² - (x+a)² - y²= 2k²;
Simplifying;
=> x² -2ax + a² + y² - x²-2ax-a²- y² =2k²
=> -4ax = 2k².
=>2ax + k² = 0.

Arithmetic:

(i) Ellipse.
(ii)

personal59:
Pls help m with this question

Double differentiation of Y with respect to X

Y= 2A sin2P Vt. Cos2P X/N


Thanks waiting for d ans in anticipation god bless

Kendzyma:
Lol,exam question num 4.. u skull in oau rit?

personal59:
Pls help m with this question

Double differentiation of Y with respect to X

Y= 2A sin2P Vt. Cos2P X/N


Thanks waiting for d ans in anticipation god bless

personal59:
Pls help m with this question

Double differentiation of Y with respect to X

Y= 2A sin2P Vt. Cos2P X/N


Thanks waiting for d ans in anticipation god bless

montty:
Hmmm I have missed

agentofchange1:


hey, my guy make your question clear , or better still , snap & post .

personal59:

We are been ask to differentiate it twice with respect to X I mean dy/dx and also second derivative d2y/dX2


The question in word cos I don't av my book here

Y = Asin2piVt.cos2pi x/n that Is A sine 2pi v t dot cos 2pi X over n op u get it am to submit by 8am pls

Kendzyma:
dy/dx=udv/dx+vdu/DX
u=sin(2πvt)
du/dx=0
v=cos(2πx/n)
dv/dx=-2π/nsin(2πx/n)
dy/dx=-2πA/nsin(2πvt)sin(2πx/n)
using product rule to find d2y/dx2
u=sin2πvt
du/dx = 0
v=sin2πx/n
dv/dx=2π/ncos2πx/n
d^2y/dx^2=-4π^2A/n^2sin(2πvt)cos(2πx/n)
d2y/dx2=-4π^2y/n^2

Richiez:
We diagnose and solve math problems here
This thread is the meeting point for nairaland math gurus...I dare anyone to ask a question in mathematics without me having an answer to them, LETS START

Kendzyma:
dy/dx=udv/dx+vdu/DX u=sin(2πvt) du/dx=0 v=cos(2πx/n) dv/dx=-2π/nsin(2πx/n) dy/dx=-2πA/nsin(2πvt)sin(2πx/n) using product rule to find d2y/dx2 u=sin2πvt du/dx = 0 v=sin2πx/n dv/dx=2π/ncos2πx/n d^2y/dx^2=-4π^2A/n^2sin(2πvt)cos(2πx/n) d2y/dx2=-4π^2y/n^2

bolaji3071:
Please help me with this question. Three times
tens digit of a two digit number is 1 less than
twice the unit digit. When the digits are
interchanged, the new number is 17 less than
twice the original number. The original number
is?
If Musa scored 75 in biology Instead of 57 his
average mark in four subjects would have been
60. What was his total marks?

bolaji3071:
Please help me with this question. Three times
tens digit of a two digit number is 1 less than
twice the unit digit. When the digits are
interchanged, the new number is 17 less than
twice the original number. The original number
is?
If Musa scored 75 in biology Instead of 57 his
average mark in four subjects would have been
60. What was his total marks?

ugonna1054:
I was getting decimal points in d first Qst! As. For the second....... Let his total score in 3 subjects be x.... Frm d first statement, a 75 score in Bio will giv him 60 average so (x + 75)/4 = 60..... x + 75 = 240.... x =165!(Score in 3 subjects)...... So his total mark will be (x + 57(real score in Bio)) = 165 + 57 =222