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3 years ago

F(x) = -1/2x^2 + 3/2x - 2 I need to find the zeros

Mathematics

1 answer:

Pachacha [2.7K]3 years ago

4 0

Answer:

{-1, 4}

Step-by-step explanation:

In the given f(x) = -1/2x^2 + 3/2x - 2

set f(x) = 0 and then multiply all the numeric terms by -2 to remove the fractions:

f(x) = -1/2x^2 + 3/2x - 2 = 0 => 1x^2 - 3x - 4 = 0

This factors into (x - 4)(x + 1) = 0, whose roots are {-1, 4}. These are the zeros.

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A person stands 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 meters

In-s [12.5K]

Answer:

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

Step-by-step explanation:

Given that,

A person stand 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 m/s.

From Pythagorean Theorem,

(The distance between car and person)²= (The distance of the car from intersection)²+ (The distance of the person from intersection)²+

Assume that the distance of the car from the intersection and from the person be x and y at any time t respectively.

∴y²= x²+10²

$\Rightarrow y=\sqrt{x^2+100}$

Differentiating with respect to t

$\frac{dy}{dt}=\frac{1}{2\sqrt{x^2+100}}. 2x\frac{dx}{dt}$

$\Rightarrow \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}. \frac{dx}{dt}$

Since the car driving towards the intersection at 13 m/s.

so, $\frac{dx}{dt}=-13$

$\therefore \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}.(-13)$

Now

$\therefore \frac{dy}{dt}|_{x=24}=\frac{24}{\sqrt{24^2+100}}.(-13)$

$=\frac{24\times (-13)}{\sqrt{676}}$

$=\frac{24\times (-13)}{26}$

= -12 m/s

Negative sign denotes the distance between the car and the person decrease.

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

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3 years ago

F(x) = -1/2x^2 + 3/2x - 2 I need to find the zeros​

F(x) = -1/2x^2 + 3/2x - 2 I need to find the zeros