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#26 Help Me ! » Triangle » 2006-12-04 07:33:41

Neha
Replies: 3

preschooltriangleyt7.gif

solve the triangle below for its unkown parts:

ok the unkown parts are the top of the triangle which is B degrees
the left side point of the triangle  is 65 degrees
right side point is C degrees

and
one side of the triangle the left side is 11inches
the other side of the triangle the right side is A inches
teh bottom of the triangle is 15inches

now i did:
65dg + C + B = 180
65dg + 65dg + B = 180
( i wrote 65dg for C angle because both sides have to be equal in my view. )
anyways
130 + B = 180
180 - 130 = 50
B = 50
C = 65
and now we have to find A side.

do like this :
11^2 + A^2 = 15^2
no right?

#27 Help Me ! » HELP OUT GUYS...plz asap thx » 2006-12-01 12:49:49

Neha
Replies: 0

[this problem has 3 questions]

Use the law of detachment or the law of sylogism to determine a conclusion that follows from statements (1) and (2)
If a valid conclusion does not follow, write no valid conclusion.

(1) if a triangle is a right triangle , then it has two acute anges.
(2) triangle ABC is a right triangle

I wrote:
p = a triangle is a right triangle
q = it has two acute angles
p --> q
law of detachment

------------------------------------------------------------------------------------------------------
(1) if three points are noncollinear,then they determine a plane
(2) points A,B,and C are coplanar

I wrote :
Law of Syllogism
a implies b, b impies c, a implies c
a -->b , b-->c, a--.c

noncollinear- points that do not lie on a single line
coplanar - lying in the same plane , any sets of three points are coplanar

--------------------------------------------------------------------------------------------------------
(1) all circles have diameters whose measure is twice the measure of the radius.
(2) AB is a radius of circle B

I wrote:
p = all circles have diameters
q = measure is twice the measure of the radius

p --> q
LAW OF DETACHMENT


( teacher gave me an x...no what can  i do )

#28 Re: Help Me ! » show that » 2006-12-01 01:48:30

oh i see thanks for the help and advice.
but if i did do it the way i did the other question it would come out to the same answer?

#29 Re: Help Me ! » show that » 2006-11-30 12:59:44

i need help on this one:

now i need help on this one......

sec^2(theta) tan(theta) - tan^3(theta)
----------------------------------------------  - cos^2(theta) = sin^2(theta)
     tan(theta)


now i did only up to here :

1  ^2     sin x          sin x  ^3
---     *  -----  -        -----
cosx       cos x         cos x                    1   ^2
---------------------------------------    -  ----      = sin^3(theta)
         sin x                                        sec
         ------
          cos x

#30 Re: Help Me ! » show that » 2006-11-30 12:56:00

rules are

cos x = 1 / sec x          sec x = 1 / cos x
sin x = 1 / csc x           csc x = 1 / sin x
tan x = sin x / cos x         cot x = cos x / sin x

--------------------------------------------------
cos^2 (x) + sin^2 (x) = 1

1 + cot^2 x = csc^2 x

#31 Help Me ! » show that » 2006-11-30 12:54:41

Neha
Replies: 5

csc^2 x - cot^2 x 
--------------------- = cos^2 x
1 + tan^2 x


csc^2 x - cot^2 x 
--------------------- = cos^2 x
1 + tan^2 x


  1               cos^2x
  -----     -    ---------
  sin^2x         sin^2x
-----------------------    =   cos^2x
   cos^2x         sin^2x
  --------   +   --------
   cos^2x         cos^2x

but 1 - cos^2 x = sin^2x, and Sin^2x + cos^2x  = 1, so the whole thing boils down to

         1
---------------   =   cos^2x
          1
      --------
       cos^2 x

#32 Help Me ! » help me out guys » 2006-11-27 04:28:25

Neha
Replies: 1

log  3^4  +  log   5^3  -  log  2^4
    3                 5                2

can you understand the way i wrote it?
i have ot simplify it

#33 Help Me ! » fred/barney » 2006-11-24 08:20:25

Neha
Replies: 1

fred can complete 5 jobs in 4 hours
when barney helps him, they can complete 9 jobs in 6 hours
how long would it take barney to complete 1 job

ok i think this is wrong but you tell me :

9/6 = 1/x
9x = 6
x = 6/9
x = 2/3
x = 0.6

help

#34 Help Me ! » hi » 2006-11-20 06:15:44

Neha
Replies: 2

find the equation of the line that passes through (2,2) and is perpendicular to the line
3y - 2x - 18  = 0

#35 Help Me ! » now what » 2006-11-16 07:36:28

Neha
Replies: 1

Which congruence test can be used to prove that triangle GHI =~ triangle LKJ?

ok                            __      __
angle G =~ angle L    GJ =~ LJ
                                __      __
angle H =~ angle K    HJ =~ KH
                                      __       __
angle GHJ =~ angle LKJ    HG =~ LK

ques3ok7.jpg


NOW WHAT.

#36 Help Me ! » horizontal » 2006-11-16 06:52:23

Neha
Replies: 1

what is the equation of a horizontal line that passes through the point at (-2,-2)?

i did
s = y2/x2 - y1/x1
and got
1.5
and my answer came out to be wrong HELP

#38 Re: Dark Discussions at Cafe Infinity » Given the opportunity... » 2006-11-15 11:25:54

i  would never live forever.
this world has too many problems and complications and hardship for a person to live in .

#39 Help Me ! » Ml » 2006-11-15 10:44:43

Neha
Replies: 1

how many ml of a 3 1/2%  salt solution must be added to 176ml of a 4 3/4% salt solution that is 45% salt?

#40 Re: Introductions » hello » 2006-11-15 06:14:51

hi soha how are you welcome to the forum. i am fine. hope you enjoy it here and get your help .
tc

#41 Re: Help Me ! » chosen » 2006-11-15 06:13:37

thank  u soooo much.

#42 Re: Help Me ! » chosen » 2006-11-15 04:55:23

guyssssssssssssssssssssss help
ok tell me do u have to multiply the numbers to get teh answer ? and then divide by 3? no right

#43 Help Me ! » chosen » 2006-11-15 04:18:55

Neha
Replies: 4

three students must be chosen to serve on a panel out of a class of 12 students . in how many different ways can they be chosen?

well 12 students in total
1 2 3 4 5 6 7 8 9 10 11 12
3 must be chosen
so it can be
123
456
789
10 11 12
or
234
567
89 10
11 12 1
or
345
678
9 10 11
12 1 2

correct
so there are  12 waysdownup

#44 Re: Help Me ! » how many » 2006-11-14 08:25:00

i dont know but that is how the problem was written....

#45 Help Me ! » how many » 2006-11-14 07:55:29

Neha
Replies: 4

how many different ways can 5 objects be arranged in a circle????????

#46 Re: Help Me ! » systems » 2006-11-11 08:09:36

I THINK YOU WRONG

a+1/2b=16
2a+b=50
2a=50-b
a=(50-b)/2
substituting in eqn. 1
(50-b)/2+1/2b=16
multiplying by 2b
b(50-b)+1=32b
50b-b^2+1-32b=0
-b^2+18b+1=0
=>b^2-18b-1=0
b=[18+/-rt(324+4)]/2
=[9+/-rt82]
similarly express b interms of a and find a
or substitute the value of a
i have a feeling your sum might be
a+(1/2)*b=16
and 2a+b=50
these are the equations of parallel lines
and hence no solution

#47 Re: Guestbook » Cn you find iloveyou. » 2006-11-10 08:45:41

oh my god i found i love you............on the third line
nice puzzle smile

#48 Re: Help Me ! » f(x) » 2006-11-10 06:49:45

oh yeah i see that now
so my answer is now -4
correct

#49 Re: Help Me ! » f(x) » 2006-11-10 06:21:26

why is it 2 instead of 1

#50 Help Me ! » f(x) » 2006-11-10 05:49:58

Neha
Replies: 4

let f(x) = x + 3  and g(x) = x^2 + 1
find: (f-g)(3)

i did>
   f(x) = x + 3
- g(x) = x^2 + 1
----------------------------
(f-g)(x) = x+3-(x^2+1)
            = x - x^2 - 2
(f-g)(3) = 3 - 3^2 - 2
            = 3 - 9 - 2
            = -8
is the answer
up

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