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Sladkaya [172]
2 years ago
5

Simplify into a binomial

Mathematics
1 answer:
Eduardwww [97]2 years ago
3 0

Answer:

x-5

Step-by-step explanation:

Consider the expression

\dfrac{3x^3-15x^2+15x-75}{3x^2+15}

Its numerator 3x^3-15x^2+15x-75 is equivalent to

3x^3-15x^2+15x-75\\ \\=3x^2(x-5)+15(x-5)\\ \\=(x-5)(3x^2+15)

Therefore, given expression can be rewritten as

\dfrac{(x-5)(3x^2+15)}{3x^2+15}=x-5

Note that 3x^2+15\neq 0, so we can divide the numerator and the denominator by non-zero expression 3x^2+15.

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Answer:

B) -4

Step-by-step explanation:

m= (0 - (-4)) / (2 - 0) = 4 / 2 = 2

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3 years ago
An object is launched at 29.4 meters per
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Answer:

The reasonable domain for the scenario is option 'a';

a) [0, 7]  

Step-by-step explanation:

For the projectile motion of the object, we are given;

The speed at which the object is launched, v = 29.4 meters per second

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f(x) = -4.9·x² + 29.4·x + 34.3

The domain for the scenario, is given by the possible values of 'x' for the function, which is found as follows;

At the height from which the object is launched, x = 0, and f(x) = 34.3

At the ground level to which the object can drop, f(x) = 0

∴ f(x) = -4.9·x² + 29.4·x + 34.3 = 0

-4.9·x² + 29.4·x + 34.3 = 0

By the quadratic formula, we have;

x = (-29.4 ± √(29.4² - 4 × (-4.9) × 34.3))/(2 × (-4.9)

∴ x = -1, or 7

Given that time is a natural number, we have the reasonable domain for the scenario as the start time when the object is launched, t = 0 to the time the object reaches the ground, t = 7

Therefore, the reasonable domain for the scenario is; 0 ≤ x ≤ 7 or [0, 7].

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