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Pavlova-9 [17]
3 years ago
14

Write the equation for the graph

Mathematics
1 answer:
bogdanovich [222]3 years ago
4 0

y=|x+1|+2

also that's a huge blank.

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How do I solve the whole thing? It doesn't make sense what so ever
Snezhnost [94]
Any value of x makes this equation true or all real numbers


-2(3x+12) is the same this as -6x-24 when distributed 
so it's -6x-24=-24-6x
her -6x just disappeared so she is wrong :)
8 0
3 years ago
Let AB be the directed line segment beginning at point A(1,1) and ending at point B(15,-3). Find the point P on the line segment
Svetllana [295]

Answer: it is 3 bc dogs are cancerous, resulting i n pythagoras himself

Step-by-step explanation:

3m/sqrt56 + 9(32x) - t = BOOMER BOOMER BOOMER BOOOMER BOOMERR

OAOAOOAAOOAOAOOAOAOAOOAOAOOAAOOAOAOA

3 0
3 years ago
Determine whether the two expressions are equivalent. Explain your reasoning.
dlinn [17]
Let us evaluate each pair of expressions one at a time.
8. (n²+4) - n² and 4n.
    (n²+4) - n² = n²+ 4 - n² = 4 which is not equal to 4n.
    NOT EQUIVALENT

9. 3x + 5 and  2(x + 3)
    2(x + 3) = 2x + 6  which is not equal to 3x + 5.
    NOT EQUIVALENT

10. 15 - 6x and  15(1 - 6x)
    15(1 - 6x) = 15 - 90x which is not equal to 15 - 6x
    NOT EQUVALENT

11. (y + y + 2 + y) + 3y and 6y + 2
    (y +y +2 + y) + 3y = y+y+y +2 + 3y = 3y + 2 + 3y = 6y + 2.
    It matches the other expression.
    EQUIVALENT

12. 8y - 3 + 10y and 3(6 - 1)
    8y - 3 + 10y = 8y + 10y - 3 = 18y - 3
    3(6 - 1) = 3(5) = 15
    The two expressions do not match.
    NOT EQUIVALENT.

Answer: Only the expressions in number 11 are equivalent.
6 0
3 years ago
Help me please I don't understand
Arturiano [62]

Step-by-step explanation:

hope this helps you out with the problems

3 0
3 years ago
Software to detect fraud in consumer phone cards tracks the number of metropolitan areas where calls originate each day. It is f
pashok25 [27]

Answer:

0.999987

Step-by-step explanation:

Given that

The user is a legitimate one = E₁

The user is a fraudulent one = E₂

The same user originates calls from two metropolitan areas  = A

Use Bay's Theorem to solve the problem

P(E₁) = 0.0131% = 0.000131

P(E₂) = 1 - P(E₁)  = 0.999869

P(A/E₁) = 3%  = 0.03

P(A/E₂) = 30% = 0.3

Given a randomly chosen user originates calls from two or more metropolitan, The probability that the user is fraudulent user is :

P(E_2/A)=\frac{P(E_2)\times P(A/E_2)}{P(E_1)\times P(A/E_1)+P(E_2)\times P(A/E_2)}

=\frac{(0.999869)(0.3)}{(0.000131)(0.03)+(0.999869)(0.3)}

\frac{0.2999607}{0.00000393+0.2999607}

\frac{0.2999607}{0.29996463}

= 0.999986898 ≈ 0.999987

6 0
3 years ago
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