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zloy xaker [14]
3 years ago
12

How do you get the x intercepts?

Mathematics
1 answer:
Marta_Voda [28]3 years ago
8 0
Factor the equation:
(x-6)(x+1)

Check:
-6+1=-5
-6*1=-6

Use the zero product property
x-6=0
x=6

x+1=0
x=-1

Final answer: x=-1, x=6
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What are the values of x and y if this equation is true? 22(x+yi)+(28+4i)+72-62i
77julia77 [94]

Answer: x=2 and y=-3

Step-by-step explanation: you’re welcome

4 0
3 years ago
Marley spins a spinner and Rose a1-6 number to the spinner has 5 equal parts numeral number 1- 5 what is the probability she wil
Taya2010 [7]
2/5 i think probably

6 0
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kozerog [31]

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hihi

Step-by-step explanation:

7 0
3 years ago
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A student is asked to find the length of the hypotenuse of a right triangle. The length of one leg is 34
VikaD [51]

Answer:

41.6 centimeters

Step-by-step explanation:

The right triangle formula: c^2 = a^2 + b^2

                                  c is the hypotenuse side of the triangle

                                  a and b are the legs of the triangle.    

 Find the hypotenuse :   c^2 = (24)^2 + (34)^2

                                          c = square root of (  (24)^2 + (34)^2)

                                          c = square root of ( 576 + 1,156)

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5 0
3 years ago
In the game of blackjack played with one​ deck, a player is initially dealt 2 different cards from the 52 different cards in the
Volgvan

Answer:

P =4.83\%

Step-by-step explanation:

First we calculate the number of possible ways to select 2 cards an ace and a card of 10 points.

There are 4 ace in the deck

There are 16 cards of 10 points in the deck

To make this calculation we use the formula of combinations

nCr=\frac{n!}{r!(n-r)!}

Where n is the total number of letters and r are chosen from them

The number of ways to choose 1 As is:

4C1 = 4

The number of ways to choose a 10-point letter is:

16C1 = 16

Therefore, the number of ways to choose an Ace and a 10-point card is:

4C1 * 16C1 = 4 * 16 = 64

Now the number of ways to choose any 2 cards from a deck of 52 cards is:

52C2 =\frac{52!}{2!(52-2)!}

52C2 = 1326

Therefore, the probability of obtaining an "blackjack" is:

P = \frac{4C1 * 16C1}{52C2}

P = \frac{64}{1326}

P = \frac{32}{663}

P =0.0483

P =4.83\%

4 0
3 years ago
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