What number does ▲ stand for in this equation? 6 × 3 = (▲ × 5)

Prove that the roots of x2+(1-k)x+k-3=0 are real for all real values of k

masha68 [24]

Answer:

Roots are not real

Step-by-step explanation:

To prove : The roots of x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0 are real for all real values of k ?

Solution :

The roots are real when discriminant is greater than equal to zero.

i.e. b^2-4ac\geq 0b

2

−4ac≥0

The quadratic equation x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0

Here, a=1, b=1-k and c=k-3

Substitute the values,

We find the discriminant,

D=(1-k)^2-4(1)(k-3)D=(1−k)

2

−4(1)(k−3)

D=1+k^2-2k-4k+12D=1+k

2

−2k−4k+12

D=k^2-6k+13D=k

2

−6k+13

D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))

For roots to be real, D ≥ 0

But the roots are imaginary therefore the roots of the given equation are not real for any value of k.

6 0

3 years ago

Which of the following statements are true? Select all that apply.

Tom [10]

Answer: The statements in options 2 and 4 are true.

Explanation:

The LCM of two numbers is the smallest number that is multiple of both numbers.

The factor of 8 and 12 are,

$8=2\times 2\times 2$

$12=2\times 2\times 3$

Since 2 and 2 are common in both but 2 and 3 are not same, therefore the L.C.M. of 8 and 12 is,

$L.C.M.(8,12)=2\times 2\times 2\times 3=24$

Therefore the statement in option 1 is incorrect.

The factor of 6 and 9 are,

$6=2\times 3$

$9=3\times 3$

Since 3 is common in both but 2 and 3 are not same, therefore the L.C.M. of 6 and 9 is,

$L.C.M.(6,9)=2\times 3\times 3=18$

Therefore the statement in option 2 is correct.

The factor of 11 and 4 are,

$11=1\times 11$

$4=2\times 2$

Since all factor are different, therefore the L.C.M. of 11 and 4 is,

$L.C.M.(11,4)=1\times 11\times 2\times 2=44$

Therefore the statement in option 3 is incorrect.

The factor of 9 and 10 are,

$9=1\times 9$

$10=2\times 5$

Since all factor are different, therefore the L.C.M. of 9 and 10 is,

$L.C.M.(9,10)=1\times 9\times 2\times 5=90$

Therefore the statement in option 4 is correct.

4 0

3 years ago

Three fair coins are tossed. If all land "heads," the player wins $10, and if exactly two land heads, the player wins $5. If it

hodyreva [135]

Answer:

After four games, a player can lose up to $ 16 to win up to $ 26. These are the probabilities for every game:

1/8 or 12.5% of landing three "heads"

3/8 or 37.5% of landing two "heads"

4/8 or 50% of landing no or only one "head".

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

If three coins land "heads" the player wins $ 10

If two coins land "heads" the player wins $ 5

Cost of playing = $ 4

2. What is the player's expected outcome after four games?

Probability of two coins out of three lands "heads" = 3/8

Probability of three coins out of three lands "heads" = 1/8

Now, let's calculate the player's expected outcome, as follows:

Four games:

Cost = 4 * 4 = $ 16

Worst-case scenario: No wins

Best-case scenario: 4 out of 4 of $ 10 win

Worst-case scenario profit or loss = 0 - 16 = Loss of $ 16

Best-case scenario profit or loss = 40 - 16 = Profit of $ 24

After four games, a player can lose up to $ 16 to win up to $ 26. These are the probabilities for every game:

1/8 or 12.5% of landing three "heads"

3/8 or 37.5% of landing two "heads"

4/8 or 50% of landing no or only one "head".

8 0

2 years ago

What are the three methods used to identify sample spaces ?

vagabundo [1.1K]

A venn diagram, a tree diagram, and a last but not least a graphical representation.

If this helps I am glad it did! (:

7 0

3 years ago

Read 2 more answers

Evaluate the following numerical expressions. 6+2^3•3

Stolb23 [73]

The answer is 30
6+8•3
6+24 =30

7 0

3 years ago

Read 2 more answers

What number does ▲ stand for in this equation? 6 × 3 = (▲ × 5) – (▲ × 2)