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Sonja [21]
4 years ago
15

There's a picture can someone help me please.

Mathematics
1 answer:
riadik2000 [5.3K]4 years ago
3 0

Answer:E) There are no real solutions

Step-by-step explanation:

Looking at the equation, x^2 +20= 4

It is a quadratic equation since the highest power of x is 2

Equation x^2 +20= 4 can be re- written as

x^2 +20= 4 = x^2 +20- 4 =0

x^2 +16= 0

x^2 + 0x +16= 0 - - - - - - - - -1

Solving with the general formula for quadratic equations,

x = [-b +/- √(b^2 -4ac)]/2a

From equation 1,

a = 1 (coefficient if x^2)

b = 0 (coefficient if x

c = 16 (value of the constant)

Substituting into the formula,

x = [-0 +/- √(0^2 -4×1×16)]/2a

= [0+/-√-64]/2×1

= +/-√-64]/2

= +/-8i/2

x= +/-4i

This is a complex number so,

There are no real solutions

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Find the smallest positive integer n such that n(n+1)(n+2) is divisible by 247.
Butoxors [25]

Answer:

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So now we need to find a number (the smallest one) that is of the form (n)(n+1)(n+2) (the product of three consecutive numbers) and that is divisible by both 13 and 19 (and therefore divisible by 247)

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Let's take a look at the multiples of 19: 19, 38, 57...

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So, if our number has both 38 and 39 as factors, then it will be divisible by 247.

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