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lyudmila [28]
2 years ago
6

Which products result in a sum or difference of cubes? Check all that apply. (x – 4)(x2 + 4x – 16) (x – 1)(x2 – x + 1) (x – 1)(x

2 + x + 1) (x + 1)( + x – 1) (x + 4)(x2 – 4x + 16) (x + 4)(x2 + 4x + 16)
Mathematics
2 answers:
olchik [2.2K]2 years ago
8 0

a^3+b^3=(a+b)(a^2-ab+b^2)\\\\a^3-b^3=(a-b)(a^2+ab+b^2)\\\\(x-4)(x^2+4x\underline{-16})\qquad \mathbb{NOT}\\\\(x-1)(x^2\underline{-x}+1)\qquad\mathbb{NOT}\\\\(x-1)(x^2+x+1)=(x-1)(x^2+(x)(1)+1^2)\qquad\mathbb{YES}\\\\(x+1)(x^2+x\underline{-1})\qquad\mathbb{NOT}\\\\(x+4)(x^2-4x+16)=(x+4)(x^2+(x)(4)+4^2)\qquad\mathbb{YES}\\\\(x+4)(x^2\underline{+4x}+4)\qquad\mathbb{NOT}

Zigmanuir [339]2 years ago
6 0

Answer:

See below.

Step-by-step explanation:

There are 2 identities:

a^3 - b^3  = (a - b)(a ^2 + ab + b^2)

a^3 + b^3 = (a + b) ( a^2 - ab + b^2)

So:-

(x - 1)(x^2 + x + 1)  = x^3 - 1    is the difference of 2 cubes.

(x + 4)(x^2 - 4x + 16) = x^3 + 64 is the sum of 2 cubes.

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What is the smallest prime number p such that
Ne4ueva [31]

The smallest prime number of p for which p^3 + 4p^2 + 4p has exactly 30 positive divisors is 43.

<h3>What is the smallest prime number of p for which p must have exactly 30 positive divisors?</h3>

The smallest number of p in the polynomial equation p^3 + 4p^2 + 4p for which p must have exactly 30 divisors can be determined by factoring the polynomial expression, then equating it to the value of 30.

i.e.

  • p^3 + 4p^2 + 4p

By factorization, we have:

  • = p(p+2)²

Now, to get exactly 30 divisor.

  • (p+2)² requires to give us 15 factors.

Therefore, we can have an equation p + 2 = p₁ × p₂²

where:

  • p₁ and  p₂ relate to different values of odd prime numbers.

So, for the least values of p + 2, Let us assume that:

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p + 2 = 5 × 3²

p + 2 = 5 × 9

p + 2 = 45

p = 45 - 2

p = 43

Therefore, we can conclude that the smallest prime number p such that

p^3 + 4p^2 + 4p has exactly 30 positive divisors is 43.

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ΔABC is a 45 - 45 - 90 triangle. The pattern of its sides is as follows:

Each leg = 1 unit (and both legs are that way, since the triangle is isosceles - so two sides are the same)

Hypotenuse = √2 units.

So if we know either leg, we multiply by √2 to get the hypotenuse. In reverse, we divide by √2 if we know the hypotenuse to get the measurement of a leg.

Our problem tells us that the hypotenuse AC is 10 units. We divide 10 by √2 to get the measurement of leg AB. Since it's a 45 -45 - 90 triangle, AB = BC.

AB = \frac{10}{\sqrt{2}}

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Thus, each leg is 5\sqrt{2} [/tex].

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