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3 years ago

22 pts PLEASE HELP ASAP JUST TELL ME THE PROPERTY NAMES I SHOULD WRITE IN THE BLANK SPACES

Mathematics

1 answer:

iogann1982 [59]3 years ago

4 0

Answer:

1. Communative Property

2. Associative Property

3. Zero Property

Step-by-step explanation:

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Step-by-step explanation:

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2 years ago

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yKpoI14uk [10]

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Four couples at a dinner party play a board game after the meal. They decide to play as teams of two and to select the teams ran

Llana [10]

Answer:

0.29

Step-by-step explanation:

We are given that there are 4 couples i.e. 8 people

They decide to play as teams of two and to select the teams randomly

Now we are asked How likely is it that every person will be teamed with someone other than the person he or she came to the party with

For people 1 , there are 6 options for pairing Since he or she cannot pair with his /her own partner

So, Choices For people 1 will be 6 person

So, Probability that person 1 will be teamed with someone other than the person he or she came to the party with = $\frac{\text{No. of choices for Person 1}}{\text{Total no. of choices available}}=\frac{6}{7}$

So, Probability that every person (=8 person) will be teamed with someone other than the person he or she came to the party with = $(\frac{6}{7})^8=0.29$

Hence Probability that every person will be teamed with someone other than the person he or she came to the party with is 0.29

4 0

3 years ago

Triangle ABC has vertices A(-5, -2), B(7, -5), and C(3, 1). Find the coordinates of the intersection of the three altitudes

Darina [25.2K]

Answer:

Orthocentre (intersection of altitudes) is at (37/10, 19/5)

Step-by-step explanation:

Given three vertices of a triangle

A(-5, -2)

B(7, -5)

C(3, 1)

Solution A by geometry

Slope AB = (yb-ya) / (xb-xa) = (-5-(-2)) / (7-(-5)) = -3/12 = -1/4

Slope of line normal to AB, nab = -1/(-1/4) = 4

Altitude of AB = line through C normal to AB

(y-yc) = nab(x-xc)

y-1 = (4)(x-3)

y = 4x-11 .........................(1)

Slope BC = (yc-yb) / (xc-yb) = (1-(-5) / (3-7)= 6 / (-4) = -3/2

Slope of line normal to BC, nbc = -1 / (-3/2) = 2/3

Altitude of BC

(y-ya) = nbc(x-xa)

y-(-2) = (2/3)(x-(-5)

y = 2x/3 + 10/3 - 2

y = (2/3)(x+2) ........................(2)

Orthocentre is at the intersection of (1) & (2)

Equate right-hand sides

4x-11 = (2/3)(x+2)

Cross multiply and simplify

12x-33 = 2x+4

10x = 37

x = 37/10 ...................(3)

substitute (3) in (2)

y = (2/3)(37/10+2)

y=(2/3)(57/10)

y = 19/5 ......................(4)

Therefore the orthocentre is at (37/10, 19/5)

Alternative Solution B using vectors

Let the position vectors of the vertices represented by

a = <-5, -2>

b = <7, -5>

c = <3, 1>

and the position vector of the orthocentre, to be found

d = <x,y>

the line perpendicular to BC through A

(a-d).(b-c) = 0 "." is the dot product

expanding

<-5-x,-2-y>.<4,-6> = 0

simplifying

6y-4x-8 = 0 ...................(5)

Similarly, line perpendicular to CA through B

<b-d>.<c-a> = 0

<7-x,-5-y>.<8,3> = 0

Expand and simplify

-3y-8x+41 = 0 ..............(6)

Solve for x, (5) + 2(6)

-20x + 74 = 0

x = 37/10 .............(7)

Substitute (7) in (6)

-3y - 8(37/10) + 41 =0

3y = 114/10

y = 19/5 .............(8)

So orthocentre is at (37/10, 19/5) as in part A.

8 0

3 years ago