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krok68 [10]
3 years ago
12

Write an equation of the graph:

Mathematics
1 answer:
Verizon [17]3 years ago
7 0

Answer:

y=\left|x+1\right|\\y =\left|x-1\right|

You can plug it into a graphing website or calculator to double check

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What are the answers to these angle questions?​
mafiozo [28]

Step-by-step explanation:

Firstly, we have to find m∠J.

Since all the angles of a Δ equal 180°, angles J, L, and K should have a sum of 180°.

So,

m∠J + m∠L + m∠K = 180°

The diagram shows us that ∠L = 49° and ∠K = 90°, so we plug in those numbers in the equation.

m∠J + 49° + 90° = 180°

Then we simplify

m∠J + 139° = 180°

Subtract 139° to both sides

∠J = 41

Now the other angles.

Since ΔJKL ~ ΔRST, then ∠J ≅ ∠R, ∠K ≅ ∠S, and ∠L ≅ ∠T

Meaning, m∠J = m∠R, m∠K = m∠S, and m∠L = m∠T

Since we know m∠J = 41°, m∠K = 90°, and m∠L = 49° we could plug those in so...

41° = m∠R , 90° = m∠S , and 49° = m∠T

4 0
3 years ago
ILL GIVE BRAINLIEST!!!!!PLEASE HELP ASAP!!!!
lana66690 [7]

Answer:

it’s pi

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
X + y = -3 & 5x - 2y = -50 ELIMINATION
Karolina [17]
I gotchu luh bro....

7 0
3 years ago
Canadians who visit the United States often buy liquor and cigarettes, which are much cheaper in the United States. However, the
victus00 [196]

Answer:

Probability of bringing a bottle of liquor into the country that is, the probability of bringing 1 bottle liquor into the country = P(B) = 0.31

The probability of not bringing a bottle of liquor into the country, that is, the probability of bringing 0 bottle liquor into the country = P(B') = 0.69

Probability distribution of bottle liquor

Let X represent the random variable of the number of bottle liquor brought into the country by a person

X | P(X)

0 | 0.69

1 | 0.31

Step-by-step explanation:

The joint probability distribution for the number of bottles of liquor and the number of cartons of cigarettes imported by Canadians who have visited the United States for 2 or more days is given in the question as

V | B

C | 0 | 1

0 | 0.62 | 0.16

1 | 0.07 | 0.15

Note that B = bottle liquor

C = Carton cigarettes

V is each variable

Let the probability of bringing a bottle of liquor into the country be P(B), that is, the probability of bringing 1 bottle liquor into the country.

The probability of not bringing a bottle of liquor into the country is P(B'), that is, the probability of bringing 0 bottle liquor into the country.

Let the probability of bringing a carton of cigarettes into the country be P(C), that is, the probability of bringing 1 carton cigarettes into the country.

The probability of not bringing a carton of cigarettes into the country is P(C'), that is, the probability of bringing 0 carton cigarettes into the country.

From the joint probability table, we can tell that

P(B n C) = 0.15

P(B n C') = 0.16

P(B' n C) = 0.07

P(B' n C') = 0.62

Find the marginal probability distribution of the number of bottles imported.

Probability of bringing a bottle of liquor into the country that is, the probability of bringing 1 bottle liquor into the country = P(B)

P(B) = P(B n C) + P(B n C') = 0.15 + 0.16 = 0.31

The probability of not bringing a bottle of liquor into the country, that is, the probability of bringing 0 bottle liquor into the country = P(B')

P(B') = P(B' n C) + P(B' n C') = 0.07 + 0.62 = 0.69

Probability distribution of bottle liquor

Let X represent the random variable of the number of bottle liquor brought into the country by a person

X | P(X)

0 | 0.69

1 | 0.31

Hope this Helps!!!

8 0
3 years ago
Plz help ill give you brainlist
charle [14.2K]
-4(x+3)=8
Using the distributive property, 
-4*x -4*(3) = 8
-4x+(-12) = 8
-4x = 8+12 = 20
x = 20/(-4) = -5
5 0
3 years ago
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