All categories

2 years ago

Which best explains why the orthocenter of an obtuse triangle is outside the triangle?

Mathematics

2 answers:

aivan3 [116]2 years ago

7 0

The answer is A. all three of the latitudes lie entirely outside the triangle

its a because all the altitudes have to cross each other so when an obtuse angle is being measured the altitudes have to cross but the only way they can can cross is if they are outside of the angle to get the right mesaurement

AlladinOne [14]2 years ago

3 0

Answer: A is the right answer.All three of the altitudes lie entirely outside the triangle.

Step-by-step explanation:

Orthocenter of a triangle is a point which formed by intersection of all the three concurrent altitudes. It can be inside or outside of a triangle. In obtuse triangle All three of the concurrent altitudes lie entirely outside the triangle .Therefore its intersection will lie outside and that intersection is orthocenter of the obtuse triangle .

You might be interested in

A 5-card hand is dealt from a perfectly shuffled deck. Define the events: A: the hand is a four of a kind (all four cards of one

TiliK225 [7]

In a hand of 5 cards, you want 4 of them to be of the same rank, and the fifth can be any of the remaining 48 cards. So if the rank of the 4-of-a-kind is fixed, there are $\binom44\binom{48}1=48$ possible hands. To account for any choice of rank, we choose 1 of the 13 possible ranks and multiply this count by $\binom{13}1=13$ . So there are 624 possible hands containing a 4-of-a-kind. Hence A occurs with probability

$\dfrac{\binom{13}1\binom44\binom{48}1}{\binom{52}5}=\dfrac{624}{2,598,960}\approx0.00024$

There are 4 aces in the deck. If exactly 1 occurs in the hand, the remaining 4 cards can be any of the remaining 48 non-ace cards, contributing $\binom41\binom{48}4=778,320$ possible hands. Exactly 2 aces are drawn in $\binom42\binom{48}3=103,776$ hands. And so on. This gives a total of

$\displaystyle\sum_{a=1}^4\binom4a\binom{48}{5-a}=886,656$

possible hands containing at least 1 ace, and hence B occurs with probability

$\dfrac{\sum\limits_{a=1}^4\binom4a\binom{48}{5-a}}{\binom{52}5}=\dfrac{18,472}{54,145}\approx0.3412$

The product of these probability is approximately 0.000082.

A and B are independent if the probability of both events occurring simultaneously is the same as the above probability, i.e. $P(A\cap B)=P(A)P(B)$ . This happens if

the hand has 4 aces and 1 non-ace, or
the hand has a non-ace 4-of-a-kind and 1 ace

The above "sub-events" are mutually exclusive and share no overlap. There are 48 possible non-aces to choose from, so the first sub-event consists of 48 possible hands. There are 12 non-ace 4-of-a-kinds and 4 choices of ace for the fifth card, so the second sub-event has a total of 12*4 = 48 possible hands. So $A\cap B$ consists of 96 possible hands, which occurs with probability

$\dfrac{96}{\binom{52}5}\approx0.0000369$

and so the events A and B are NOT independent.

4 0

3 years ago

Now You Try!

tresset_1 [31]

Answer:

Step-by-step explanation:

6 0

2 years ago

Type the correct answer in the box. Round your answer to the hundredth.

sesenic [268]

Answer:

Approximately 22.97 years

Step-by-step explanation:

Use the equation for continuously compounded interest, which uses the exponential base "e":

$A=P e^{k*t}$

Where P is the principal (initial amount of the deposit - unknown in our case)

A is the accrued value (value accumulated after interest is compounded), in our case it is not a given value but we know that it triples the original deposit (principal) so we write it as: 3 P (three times the principal)

k is the interest rate : 5% which translates into 0.05

and t is the time in the savings account to triple its value (what we need to find)

The formula becomes:

$3P = P e^{0.05 * t}$

To solve for "t" we divide both sides of the equation by P (notice it cancels P everywhere), and then to solve for the exponent "t" we use the natural logarithm function:

$\frac{3P}{P} = \frac{P}{P} e^{0.05 * t}$