Find the distance between (–6, 5) and (3, 0). round to the nearest tenth if necessary. a 106.1 b

A triangle has vertices at (1,10), (-5,2), and (7,2). What is its orthocenter. Show your work.

kondor19780726 [428]

Answer:

Question: What is the orthocenter of a triangle with the vertices (-1,2) (5,2) and (2,1)?

The coordinates of point A are (-1,2), point B are (5,2), and point C are (2,1).

The orthocent is the intersection of the three altitudes. An altitude goes from a vertex and is perpendicular to the line containing the opposite side.

In the coordinate plane the equations of the altitudes can be found and then a system of equations can be solved.

Altitude 1. From point C perpendicular to the line containing side AB.

Slope of line AB is 0 (horizontal line), a vertical line is perpendicular to a horizontal line. Thus, the equation of altitude 1 is x=2 .

Altitude 2. From point B perpendicular to the line containing side AC.

Slope of line AC is −13 , the slope of a line perpendicular to line AC is 3. The equation of altitude 2 is y=3x−13

Altitude 3. From point A perpendicular to the line containing side BC.

Slope of line BC is 13 , the slope of a line perpendicular to line BC is −3 . The equation of altitude 3 is y=−3x−1

The orthocenter is the point where all three altitudes intersect.

x=2

y=3x−13

y=−3x−1

Use substitution to solve the first two equations y=3(2)−13=−7

The orthocenter is the point (2,−7)

we did not need the third equation, but we can use it as a check, plug the coordinates into the third equation:

−7=−3(2)−1

−7=−6−1

−7=−7 it works.

3 0

3 years ago

There are two calculus classes at your school classes have a class average of 75.5 the first class as a standard deviation of 10

DENIUS [597]

To compare the two classes, the Coefficient of Variation (COV) can be used.

The formula for COV is this:
C = s / x

where s is the standard deviation and x is the mean

For the first class:
C1 = 10.2 / 75.5
C1 = 0.1351 (13.51%)

For the second class:
C2 = 22.5 / 75.5
C2 = 0.2980 (29.80%)

The COV is a test of homogeneity. Looking at the values, the first class has more students having a grade closer to the average than the second class.

8 0

3 years ago

What is the length of AC¯?

liberstina [14]

X+4=-4
3x-8=-8

Hope it helps.

4 0

3 years ago

Read 2 more answers

In this exercise, we consider strings made from uppercase letters in the English alphabet and decimal digits.

Yuri [45]

Answer:

a) $26^2 36^8$

b) $21\cdot10\cdot36^7$

c) $5^3 31^7$

d) $10\cdot 9\cdot 8 \cdot 7 \cdot 26^6$

Step-by-step explanation:

We will use the product rule from combinatorics.

a) There are 26 letters in the English alphabet, so there are 26 possible choices for the first character and 26 possible choices for the last one. Each one of the remaining eight characters of the string has 36 choices (letters or digits). By the product rule, there are $26\cdot36\cdot 36\cdots 36\cdot 26=26^2 36^8$ strings.

b) We have 5 possible choices for the first character, it must be some vowel a,e,i,o,u. The second character can be chosen in 21 ways, selecting some consonant. There are 10 possibilities for the last character because only of the digits are allowed. The other seven characters have no restrictions, so each one can be chosen in 36 ways. By the product rule there are $21\cdot 10\cdot 36^7$ strings.

c) The third character has 5 possibilities. Repetition of vowels is allowed, so the sixth and eighth characters have each one 5 possible choices. There are seven characters left. None of them are a vowel, but they are allowed to take any other letter or digit, so each one of them can be chosen in 36-5=31 ways. Therefore there are $5^3 31^7$ strings.

d) Remember that the binomial coefficient $\binom{n}{k}$ is the number of ways of choosing k elements from a set of n elements. In this case, to count all the possible strings, we first need to count in how many ways we can select the four positions that will have the digits. This can be done in $\binom{10}{4}$ ways, since we are choosing four elements from the set of the ten positions of the string. Now, for the first position, we can choose any digit so it has 10 possibilities. The second position has 9 possibilities, because we can't repeat the digit used on the first position. Similarly, there are 8 choices for the third position and there are 7 choices for the fourth. Now, these are the only digits on the string, so the remaining 6 characters must be letters, then each one of them has 26 possibilities. By the product rule, there are $10\cdot 9\cdot 8 \cdot 7 \cdot 26^6$ strings.

3 0

3 years ago

Five squares are positioned as shown. The small square indicated has area 1 . What is the value of h?

shepuryov [24]

Answer:

4

Step-by-step explanation:

Each box is 1 larger than the earlier one, so the boxes are 1, 2, 3, 4, and 5.

4 + 5 is the width across the bottom

2 + 3 is the width across the top (not counting h)

9 - 5 = 4, so h = 4

4 0

2 years ago

Find the distance between (–6, 5) and (3, 0). round to the nearest tenth if necessary. a 106.1 b –10.3 c 10.3 d 5.8