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3 years ago

Translate the following " Three times a number is at least that number plus 1"

Mathematics

2 answers:

Zielflug [23.3K]3 years ago

6 0

Answer:

$3n \geq n + 1$

Step-by-step explanation:

3 times a number = 3n

is at least = greater than or equal to, so $\geq$

that number plus 1 = n + 1

the same variable is used (n) because the number is supposed to be the same.

Put that all together and u get $3n \geq n + 1$

hope this helps!

Yuri [45]3 years ago

4 0

Answer:

3 x 2+1

Step-by-step explanation:

I think sorry kung mali

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Answer:

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Step-by-step explanation:

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Answer and Step-by-step explanation:

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Step-by-step explanation:

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2 years ago

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Basile [38]

Answer:

$f(x) = x^3 - 2x^2 -3x + 6$

Step-by-step explanation:

According to the Factor Theorem, if (x - k) is a factor of a polynomial P(x), then P(k) must equal zero.

We are given that a polynomial function has the zeros 2, √3, and -√3. So, we can let k = 2, √3, -√3.

So, according to the Factor Theorem, P(2), P(√3) and P(-√3) must equal 0.

Testing each choice, we can see that only A is true:

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Testing all three values yields that:

$\displaystyle \begin{aligned} f(2) &= (2)^3 - 2(2)^2 -3(2) + 6 \\ &= (8) - (8) -(6) + (6) \\ &= 0\stackrel{\checkmark}{=}0 \\ \displaystyle f(\sqrt{3}) &= (\sqrt{3})^3 - 2(\sqrt{3})^2 - 3(\sqrt{3}) + 6 \\ &=(3\sqrt{3}) -(6)-(3\sqrt{3}) + 6 \\ &= 0\stackrel{\checkmark}{=}0 \\ f(-\sqrt{3}) &= (-\sqrt{3})^3 - 2(-\sqrt{3})^2 - 3(-\sqrt{3}) + 6 \\ &=(-3\sqrt{3}) -(6)+(3\sqrt{3}) + 6 \\ &= 0\stackrel{\checkmark}{=}0 \end{aligned}$

Hence, our answer is A.

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Translate the following " Three times a number is at least that number plus 1"​

Translate the following " Three times a number is at least that number plus 1"