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DIA [1.3K]
3 years ago
12

This is the question i need help with

Mathematics
1 answer:
vekshin13 years ago
4 0
I'm pretty sure it should be the 45. or A
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What is the value of the discriminant in the equation shown below?<br><br> x^2 -6x +4 =0
irakobra [83]

Answer:

Discriminant = 20

Step-by-step explanation:

We use formula to find the discriminant.

Discriminant (D) = b^2 - 4ac

The given equation is x^2 - 6x + 4 = 0.

Here the value of a = 1, b = -6 and c = 4.

Plug in the given values in the formula, we get

Discriminant (D) = (-6)^2 - 4*1*4

= 36 - 16

Discriminant = 20

Thank you.

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3 years ago
Find the general solution to "+′−6=3^2
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Answer:

answer is -6=9

Step-by-step explanation:

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Step-by-step explanation:

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2 years ago
Which summation formula represents the series below? 1 + 2 + 6 + 24
krek1111 [17]

Question:

Which summation formula represents the series below? 1 + 2 + 6 + 24

(a) \sum_{n=2}^{5}(n-1) !

(b) \sum_{n=0}^{3} n !

(c) \sum_{n=1}^{4}(n+1) !

(d) \sum_{n=2}^{5} n !

Answer:

Option a: \sum_{n=2}^{5}(n-1) ! is the correct answer.

Explanation:

Option a: \sum_{n=2}^{5}(n-1) !

By substituting the values of n and expanding the summation, we have,

(2-1) !+(3-1) !+(4-1) !+(5-1) !

Subtracting, we have,

1 !+2!+3 !+4 !

Expanding the factorial,

1+(2*1)+(3*2*1)+(4*3*2*1)

Simplifying, we get,

1+2+6+24

Thus, the summation \sum_{n=2}^{5}(n-1) ! represents the series 1+2+6+24

Hence, Option a is the correct answer.

Option b: \sum_{n=0}^{3} n !

By substituting the values of n and expanding the summation, we have,

0!+1!+2!+3!

Expanding the factorial,

0+1+(2*1)+(3*2*1)

Simplifying, we get,

0+1+2+6

Thus, the summation \sum_{n=0}^{3} n ! does not represents the series 1+2+6+24

Hence, Option b is not the correct answer.

Option c: \sum_{n=1}^{4}(n+1) !

By substituting the values of n and expanding the summation, we have,

(1+1) !+(2+1) !+(3+1) !+(4+1) !

Adding, we have,

2!+3!+4!+5!

Expanding the factorial,

(2*1)+(3*2*1)+(4*3*2*1)+(5*4*3*2*1)

Simplifying, we get,

2+6+24+120

Thus, the summation \sum_{n=1}^{4}(n+1) ! does not represents the series 1+2+6+24

Hence, Option c is not the correct answer.

Option d: \sum_{n=2}^{5} n !

By substituting the values of n and expanding the summation, we have,

2!+3!+4!+5!

Expanding the factorial,

(2*1)+(3*2*1)+(4*3*2*1)+(5*4*3*2*1)

Simplifying, we get,

2+6+24+120

Thus, the summation \sum_{n=2}^{5} n ! does not represents the series 1+2+6+24

Hence, Option d is not the correct answer.

Hence, the correct answer is Option a: \sum_{n=2}^{5}(n-1) !

6 0
3 years ago
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