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Ymorist [56]
2 years ago
9

Here is a tree diagram showing the sample space for two independent events. How many outcomes are there for the second event?

Mathematics
1 answer:
kobusy [5.1K]2 years ago
8 0

Answer:

probably 8 just remember

________________________________________________________

sick of crying  

tired of trying

yes i'm smiling

but inside i'm  

       dying

Step-by-step explanation:

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Factor the expression x-x^2y^2
ivann1987 [24]

Factor the expression x-x^2y

5 0
3 years ago
1/10 of 3000 is 10 times as much as
-BARSIC- [3]
1/10 x 3000 (of means times) = 300
300 is 10 times more than what ur looking for
300/10=30
300 is 10 time more than 30
3 0
4 years ago
Read 2 more answers
In each diagram, line k is parallel to line l, and line t intersects line k and l.
viva [34]

Answer:

D) the value of x is 135, because the angles shown in each diagram are complementary​

Step-by-step explanation:

Each diagram has 2 angles labeled, and those two angles are always supplementary. The sum of the measures of two supplementary angles is 180°.

x + 45 = 180

x = 180 - 45

x = 135

Answer: D) the value of x is 135, because the angles shown in each diagram are complementary​

4 0
3 years ago
Which equation has solutions of 6 and -6? x2 – 12x + 36 = 0 x2 + 12x – 36 = 0 x2 + 36 = 0 x2 – 36 = 0
Svetlanka [38]

<u>Answer:</u>

The equation that has solutions 6 and -6 is x^2 - 36 = 0

<u>Solution:</u>

We have to find which equation has the solutions 6 and -6.

We have been given three equations.

x^{2}-12 x+36=0  --- eqn 1

x^{2}+12 x-36=0 -- eqn 2

x^{2}-36=0  ---- eqn 3

The 6 and -6 to satisfy any of these equations they have to be the roots of the equation.

This means that when we substitute 6 and -6 in any of the equations and then solve them the answer on simplification should be 0.

This condition should individually be satisfied by both 6 and -6 for any one of the equations.

Now let us try and substitute 6 and -6 in eq1.

Now, substituting 6 in eq1.

62-12×6+36=0

Now we simply the equation to check is the LHS is equal to the RHS of the equation.  

LHS:

72-72=0

RHS:  0  

Since LHS=RHS it is the root of the equation.

Now we check if -6 satisfies eq1.

-62-12×-6+36=0

LHS:

72+72=144

RHS:  0

Hence LHS is not equal to RHS, -6 is not the root of eq1.

Similarly we check for eq2  

Checking for 6 and -6 we get

LHS is not equal to RHS hence this does not satisfy eq2.

Now in the same way we check for eq3

LHS=RHS for both 6 and -6 hence they are the solutions for eq3.

Hence the equation that has solutions 6 and -6 is x^2 - 36 = 0

3 0
3 years ago
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Halp me with these down belowww
Simora [160]
For the first image it is option 4
Second image it is option 1
third image it is 3 dogs
3 0
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