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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?
[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 )
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?
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
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
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
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
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
find the equation of the line that passes through (2,2) and is perpendicular to the line
3y - 2x - 18 = 0
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
NOW WHAT.
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
you dont believe in life after death???
i would never live forever.
this world has too many problems and complications and hardship for a person to live in .
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?
hi soha how are you welcome to the forum. i am fine. hope you enjoy it here and get your help .
tc
thank u soooo much.
guyssssssssssssssssssssss help
ok tell me do u have to multiply the numbers to get teh answer ? and then divide by 3? no right
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 ways
i dont know but that is how the problem was written....
how many different ways can 5 objects be arranged in a circle????????
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
oh my god i found i love you............on the third line
nice puzzle
oh yeah i see that now
so my answer is now -4
correct
why is it 2 instead of 1
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