True or false the area of a regular heptagon can be found by breaking the heptagon into seven congruent triangles and then takin
g the sum of their areas
2 answers:
Answer: The answer is true.
Step-by-step explanation: The given statement is :
" The area of a regular heptagon can be found by breaking the heptagon into seven congruent triangles and then taking the sum of their areas".
We are to check its is true or false.
In the attached figure, a regular heptagon ABCDEFG is drawn. Sine the heptagon is regular, so all its sides are equal and all its angles are also equal. So, we can break the heptagon into seven congruent triangles 1, 2, 3, 4, 5, 6 and 7 taking 'O' as the centre of the heptagon.
Finding the areas of these seven triangles and taking the sum will result in the area of the heptagon ABCDEFG.
Thus, the given statement is "TRUE".
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Yay, easy peasy, nice refresher
53. evaluate for x=infinity
we get 100/(infiniy^2-5) which is basically 100/infinity which is a very small number so about 0
54.
vertical assemtotes, neato
that is where the denomenator equals 0
but wait, simplify first
ok, fraction is simplified (anything factored out would be a hole)
so set the denom to 0 and solve
x^2-5=0
(x-√5)(x+√5)=0
x=-√5 and √5
vertical assemtotes at x=-√5 and x=√5
55. horizontal assemtotes
for the function p(x)/f(x)
if p(x) has lower degree than f(x) then y=0 is the horizontal assemtote
if p(x) has equal degree to f(x) then divide the leading coefients of p(x) and f(x)
if p(x) has greater degree than f(x) then we got slant assemtotes
100/(x^2-5)
100 has lower degree than x^2
y=0 is the horizontal assemtote
56. evalate for x=-5
we get -150/-10=150/10=15
the limit is 15
57. that negative on the right side means that we must approach it from the left
let's see if we can factor out the denomenator
nope
so go and evaluate values like 4, 4.5, 4.9, etc, approaching x=5 from the left
we keep getting smaller and smaller valuse
this make sense because as x approaches 5, we get 30(c)/(c-5) where c is close to 5, and as c appraoches 5 very close, the top part is appraoching 150 but the bottom approaches a very small negative number, and if we do something like 150/-0.00000000001, we get a very negative number
the limit is negative infinity
alternately, you could figure that there is a vertical assemtote at x=5 so we can see that if we aproximaate closer and closer values, we get smaller values. so from what we know about assemtotes, expecially vertical ones, we say that it approaches negative infinity
58.
evalauating we get infintity/inifinty, not helpful
hmmm, is there a horizontal assemtote to help us?
the degree of top is higher so we got a slant assemtote
the only way is to use long division to divide the top by the bottom (see attachment)
we get
the slant assemtote is
as you can see, as x approaches infinity, y also will approach infinity
ANSWERS:
53. 0
54. VA is x=-√5 and x=√5
55. HA is y=0
56. the limit is 15
57. the limit is -∞
58. the limit is ∞
53. evaluate for x=infinity
we get 100/(infiniy^2-5) which is basically 100/infinity which is a very small number so about 0
54.
vertical assemtotes, neato
that is where the denomenator equals 0
but wait, simplify first
ok, fraction is simplified (anything factored out would be a hole)
so set the denom to 0 and solve
x^2-5=0
(x-√5)(x+√5)=0
x=-√5 and √5
vertical assemtotes at x=-√5 and x=√5
55. horizontal assemtotes
for the function p(x)/f(x)
if p(x) has lower degree than f(x) then y=0 is the horizontal assemtote
if p(x) has equal degree to f(x) then divide the leading coefients of p(x) and f(x)
if p(x) has greater degree than f(x) then we got slant assemtotes
100/(x^2-5)
100 has lower degree than x^2
y=0 is the horizontal assemtote
56. evalate for x=-5
we get -150/-10=150/10=15
the limit is 15
57. that negative on the right side means that we must approach it from the left
let's see if we can factor out the denomenator
nope
so go and evaluate values like 4, 4.5, 4.9, etc, approaching x=5 from the left
we keep getting smaller and smaller valuse
this make sense because as x approaches 5, we get 30(c)/(c-5) where c is close to 5, and as c appraoches 5 very close, the top part is appraoching 150 but the bottom approaches a very small negative number, and if we do something like 150/-0.00000000001, we get a very negative number
the limit is negative infinity
alternately, you could figure that there is a vertical assemtote at x=5 so we can see that if we aproximaate closer and closer values, we get smaller values. so from what we know about assemtotes, expecially vertical ones, we say that it approaches negative infinity
58.
evalauating we get infintity/inifinty, not helpful
hmmm, is there a horizontal assemtote to help us?
the degree of top is higher so we got a slant assemtote
the only way is to use long division to divide the top by the bottom (see attachment)
we get
the slant assemtote is
as you can see, as x approaches infinity, y also will approach infinity
ANSWERS:
53. 0
54. VA is x=-√5 and x=√5
55. HA is y=0
56. the limit is 15
57. the limit is -∞
58. the limit is ∞