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8_murik_8 [283]
3 years ago
10

The quotient of 3 times a number "x" and 2 less 3 times the square of "x"​

Mathematics
1 answer:
Fofino [41]3 years ago
6 0

Three times a number x can be written as

3x

The quotien between this and 2 is

\dfrac{3x}{2}

from this, we will subtract something:

\dfrac{3x}{2}-\ldots

And this "something" is three times the square of x. The square of x is x^2, and three times this is 3x^2

So, our final expression is

\dfrac{3x}{2}-3x^2

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Can someone help asap!!
Law Incorporation [45]
Just look up a calculator of that type of work on your phone and it will help I promise
4 0
3 years ago
The function f(x) = –x2 – 4x + 5 is shown on the graph.
Masja [62]
I believe it’s The domain of the function is all real numbers less than or equal to -2
6 0
3 years ago
Sketch the situation if necessary and use related rates to solve. Two airplanes are flying in the air at the same height. Airpla
Alexeev081 [22]

Answer:

  -390 mph

Step-by-step explanation:

Let a and b represent, respectively, the distances of A and B from the airport. The distance d between the planes is then given by the Pythagorean theorem as ...

  d² = a² + b²

Differentiating with respect to time, we have ...

  2d·d' = 2a·a' +2b·b'

Solving for d', we get ...

  d' = (a/d)a' +(b/d)b'

The value of d at the time of interest is ...

  d = √(a² +b²) = √(30² +40²) = √2500 = 50

Then the rate of change of separation is ...

  d' = (30/50)(-250 mph) +(40/50)(-300 mph) = (-150 -240) mph

  d' = -390 mph

The distance between planes is decreasing at 390 miles per hour.

4 0
3 years ago
Is he correct and show work please
mario62 [17]

9514 1404 393

Answer:

  yes

Step-by-step explanation:

f(0) = -3 +0.5×0 = -3

f(1) = -3 +0.5×1 = -2.5 = f(0) +0.5

Two linear functions are the same if they agree at two points. Here, f(0) and f(1) are the same for both forms of the function, so these different forms represent the same relation.

Diego is correct.

5 0
3 years ago
Identify the zeros of the function f(x) = 2x^2 − 4x + 5 using the Quadratic Formula
Nostrana [21]

For this case we have that by definition, the roots, or also called zeros, of the quadratic function are those values of x for which the expression is 0.

Then, we must find the roots of:

2x ^ 2-4x + 5 = 0

Where:

a = 2\\b = -4\\c = 5

We have to:x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}

Substituting we have:

x = \frac {- (- 4) \pm \sqrt {(- 4) ^ 2-4 (2) (5)}} {2 (2)}\\x = \frac {4 \pm \sqrt {16-40}} {4}\\x = \frac {4 \pm \sqrt {-24}} {4}

By definition we have to:

i ^ 2 = -1

So:

x = \frac {4 \pm \sqrt {24i ^ 2}} {4}\\x = \frac {4 \pm i \sqrt {24}} {4}\\x = \frac {4 \pm i \sqrt {2 ^ 2 * 6}} {4}\\x = \frac {4 \pm 2i \sqrt {6}} {4}\\x = \frac {2 \pm i \sqrt {6}} {2}

Thus, we have two roots:

x_ {1} = \frac {2 + i \sqrt {6}} {2}\\x_ {2} = \frac {2-i \sqrt {6}} {2}

Answer:

x_ {1} = \frac {2 + i \sqrt {6}} {2}\\x_ {2} = \frac {2-i \sqrt {6}} {2}

3 0
4 years ago
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