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mel-nik [20]
3 years ago
14

Can I get help on finding the volume of a cone ? :/

Mathematics
2 answers:
Assoli18 [71]3 years ago
7 0
It is 803.84in2 is correct
Anna71 [15]3 years ago
3 0
I believe that the answer is 803.84 in².
You might be interested in
Find two consecutive integers whose product is 50
GarryVolchara [31]

n, n+1 - two consecutive integers

n(n + 1) = 50     <em>use distributive property</em>

n² + n = 50     <em>subtract 50 from both sides</em>

n² + n - 50 = 0

-----------------------------------------------------

ax² + bx + c =0

if b² - 4ac > 0 then we have two solutions:

[-b - √(b² - 4ac)]/2a and [-b - √(b² + 4ac)]/2a

if b² - 4ac = 0 then we have one solution -b/2a

if  b² - 4ac < 0 then no real solution

----------------------------------------------------------

n² + n - 50 = 0

a = 1, b = 1, c = -50

b² - 4ac = 1² - 4(1)(-50) = 1 + 200 = 201 > 0 → two solutions

√(b² - 4ac) = √(201) - it's the irrational number

Answer: There are no two consecutive integers whose product is 50.

3 0
3 years ago
ANSWER PLZ QUICK AND FAST WHAT THE ANSWER QUICK AND FAST WHAT THE ANSWER QUICK
WARRIOR [948]
Hi there!

Let's have a look at all the symbols.

< means "smaller than".

\geqslant means "greater than or equal to".

\leqslant means "smaller than or equal to".

> means "greater than".

Hence, the answer is C.
~ Hope this helps you!
4 0
3 years ago
Read 2 more answers
I need help please ​
just olya [345]

Answer:

a?

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
What is the slope of the line that contains the following 2 points (1,2) (4, 0)
Eduardwww [97]
You use the rule , y2-y1/x2-x1 , to find the slope so the answer will be = 0-2/4-1 = -2/3
5 0
2 years ago
(a) If G is a finite group of even order, show that there must be an element a = e, such that a−1 = a (b) Give an example to sho
Dahasolnce [82]

Answer:

See proof below

Step-by-step explanation:

First, notice that if a≠e and a^-1=a, then a²=e (this is an equivalent way of formulating the problem).

a) Since G has even order, |G|=2n for some positive number n. Let e be the identity element of G. Then A=G\{e} is a set with 2n-1 elements.

Now reason inductively with A by "pairing elements with its inverses":

List A as A={a1,a2,a3,...,a_(2n-1)}. If a1²=e, then we have proved the theorem.

If not, then a1^(-1)≠a1, hence a1^(-1)=aj for some j>1 (it is impossible that a^(-1)=e, since e is the only element in G such that e^(-1)=e). Reorder the elements of A in such a way that a2=a^(-1), therefore a2^(-1)=a1.

Now consider the set A\{a1,a2}={a3,a4,...,a_(2n-1)}. If a3²=e, then we have proved the theorem.

If not, then a3^(-1)≠a1, hence we can reorder this set to get a3^(-1)=a4 (it is impossible that a^(-1)∈{e,a1,a2} because inverses are unique and e^(-1)=e, a1^(-1)=a2, a2^(-1)=a1 and a3∉{e,a1,a2}.

Again, consider A\{a1,a2,a3,a4}={a5,a6,...,a_(2n-1)} and repeat this reasoning. In the k-th step, either we proved the theorem, or obtained that a_(2k-1)^(-1)=a_(2k)

After n-1 steps, if the theorem has not been proven, we end up with the set A\{a1,a2,a3,a4,...,a_(2n-3), a_(2n-2)}={a_(2n-1)}. By process of elimination, we must have that a_(2n-1)^(-1)=a_(2n-1), since this last element was not chosen from any of the previous inverses. Additionally, a_(2n1)≠e by construction. Hence, in any case, the statement holds true.

b) Consider the group (Z3,+), the integers modulo 3 with addition modulo 3. (Z3={0,1,2}). Z3 has odd order, namely |Z3|=3.

Here, e=0. Note that 1²=1+1=2≠e, and 2²=2+2=4mod3=1≠e. Therefore the conclusion of part a) does not hold

7 0
3 years ago
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