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Hello every one, how are you?

Convert 0.75 in base 10 to a number in base 5. Kindly help out! ASAP

MhizVeronica:
wow great one op, i really learn a lot from this


i rep funaab

What level and deptmt?


Am Mathematics set 2014

Pls d Gurus, kindly state d best mathematics textbook with both pure/further maths to buy because I need one urgent for secondary school student.

ladokuntlad:
What level and deptmt?


Am Mathematics set 2014

100l mathematics too

Trone:
Convert 0.75 in base 10 to a number in base 5. Kindly help out! ASAP

Convert 0.75 (base 10) to base 5 :
Algorithm :
(1) Multiply number in base 10 by 5 : 0.75 *
5 = 3.75

(2) Separate integer from fraction : 3.75 = 3
+ 0.75

(3) Integer is the first place in the decimal
of the base 5 number, i.e., decimal so far is
0.3.

(4) Go back to (1) and multiply decimal just
obtained by 5 again.

In this case, we'll just be repeating the
same thing (0.75 *5 = 3.75) over and over.
Thus, the decimal expansion is 0.333

NB: This is a rare case were the decimal part gives a repeated fraction. Most of this type of questions comes much in handy... See goggle

Summary
The key idea of converting a fraction (i.e., a
value with a finite number of digits right of the
radix point) is to multiply repeatedly by K, and
determine if the result is bigger than 1 or not. If
so, store the integer part in an array. If not, store
0 in the array.
Then, subtract off the integer part. Repeat until
the number is 0. If the number is 0, it may
simply be a repeated fraction.

2nioshine:
Convert 0.75 (base 10) to base 5 :
Algorithm :
(1) Multiply number in base 10 by 5 : 0.75 *
5 = 3.75

(2) Separate integer from fraction : 3.75 = 3
+ 0.75

(3) Integer is the first place in the decimal
of the base 5 number, i.e., decimal so far is
0.3.

(4) Go back to (1) and multiply decimal just
obtained by 5 again.

In this case, we'll just be repeating the
same thing (0.75 *5 = 3.75) over and over.
Thus, the decimal expansion is 0.333

NB: This is a rare case were the decimal part gives a repeated fraction. Most of this type of questions comes much in handy... See goggle

Summary
The key idea of converting a fraction (i.e., a
value with a finite number of digits right of the
radix point) is to multiply repeatedly by K, and
determine if the result is bigger than 1 or not. If
so, store the integer part in an array. If not, store
0 in the array.
Then, subtract off the integer part. Repeat until
the number is 0. If the number is 0, it may
simply be a repeated fraction.

Thank you!

MhizVeronica:
100l mathematics too

Stil av a long way to go.


Mathematics is fun. Get mentally prepared for abstract reasoning

@moderator and d likes.

Why was my solution hidden?

Trone:
Convert 0.75 in base 10 to a number in base 5. Kindly help out! ASAP

I gave a solution to this particular problem about 2-3days now. And on checking tonight, it's nowhere to be found

Aksimple47:
Pls d Gurus, kindly state d best mathematics textbook with both pure/further maths to buy because I need one urgent for secondary school student.

Project 1

Scholastic Series 2

Project 3

Engineering Maths by K.A Stroud of for very advanced level, engineering maths by H.K. Dass.

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ladokuntlad:
Stil av a long way to go.


Mathematics is fun. Get mentally prepared for abstract reasoning

drop ur whatsapp or fb id pls

MhizVeronica:
drop ur whatsapp or fb id pls

My WhatsApp number is on my signature while my fb id is ladokun olatunji

smiles. ......... nairaland maths clinic..... long time old friends....2nioshine,ladokuntlad and others..... thanx for keeping the thread alive @calculusf (x)

Please help with this.
1) find an equation of the tangent line to the curve y=tan^2 x at the point ((pi÷3),3)

2) find integral of cos ^6 x DX

3) if xy + y^2 =1 find (d^2 y ÷ dx^2) at (0,1)

Thanks

D

kingmayak:
Please help with this.
1) find an equation of the tangent line to the curve y=tan^2 x at the point ((pi÷3),3)

2) find integral of cos ^6 x DX

3) if xy + y^2 =1 find (d^2 y ÷ dx^2) at (0,1)

Thanks

here

Thanks a lot buh (3) didn't see it clearly.

kingmayak:
Thanks a lot buh (3) didn't see it clearly.

ok leme type it out

Given that. xy+y² =1

getting y" at (0,1)

first differentiating implicitly

=> y+xy' +2yy' =0 .......(*)

at (0,1) . => 1 +0(y') +2y'=0

y'= -1/2


again , from (*)

y'+ y' +xy"+2(y')² +2yy" =0

at (0,1) we have

2(-1/2) +0(y" ) + 2(-1/2)² +2(1)y"=0

thus y"= 1/4


thats it

hope you get ?

Thanks alot

kingmayak:
Thanks alot

uwc buddy...

Pls guys help on these questions 1. Find an equation of the tangent line to the curve y=tan^2 x. At the point (pi/3,3)
(2).find the integral of cos^6 xdx
(3).if xy+y^2=1.find d^2y/dx^2 at (0,1).

Soneh:
Pls guys help on these questions 1. Find an equation of the tangent line to the curve y=tan^2 x. At the point (pi/3,3)
(2).find the integral of cos^6 xdx
(3).if xy+y^2=1.find d^2y/dx^2 at (0,1).

scroll up . its solved bro.

Please I need help with this question. I just can't get around it. The Circle Geometry theorems I know don't seem to work.

The question is, Find x, i.e. AB.

@ agentofchange is dat nt numerical analysis qustion frm HK DASS

oluwasammy:
@ agentofchange is dat nt numerical analysis qustion frm HK DASS

what. question. sir ?

Good day to you all. It's been quite a while. I have been on a huge research concerning the order of operations in Mathematics but I am back and better now. How have you guys been faring?

Cortana2:
Please I need help with this question. I just can't get around it. The Circle Geometry theorems I know don't seem to work.

The question is, Find x, i.e. AB.

Please I still need help with this question.

Cortana2:
Please I still need help with this question.

Maybe you should try this theorem

If a chord and a tangent intersect
externally, then the product of the
length of segments of the chord is
equal to the square of the length of
the tangent from the point of contact
to the point of intersection


Thus AB*BC =(CT) ^2

6x=8^2

And this gives

x=64/6=10.67

Try and do a cross check it's been long i worked on circle geometry

Cortana2:
Please I need help with this question. I just can't get around it. The Circle Geometry theorems I know don't seem to work.

The question is, Find x, i.e. AB.

The correct theorem states that,"If a secant and a tangent of a circle are drawn from a point outside the circle, then the product of the lengths of the secant and its external segment equals the square of the length of the tangent segment. Interpreted as AC × BC = CT²
(6+x)6 = 8² = 64
6 + x = 64 ÷ 6 = 10.67
x = 10.67 - 6
x = 4.67