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Airida [17]
3 years ago
7

If the domain of the function f(x) = x - 5 is {1, 3, 4, 6} what is the range? F

Mathematics
1 answer:
Margarita [4]3 years ago
7 0

Answer:

-4, -2, -1, 1

Step-by-step explanation:

If the domain is given we use the equation to determine the range but usually the range would be everything in this equation.

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Point B is on line segment AC . Given AB=3x,AB=3x, BC=4x+8,BC=4x+8, and AC=5x+10,AC=5x+10, determine the numerical length of AC
kolezko [41]

Answer:

AC = 15

Step-by-step explanation:

3x + 4x + 8 = 5x + 10

7x + 8 = 5x + 10

2x + 8 = 10

2x = 2

x = 1

5(1) + 10 = 15

3 0
3 years ago
Read 2 more answers
1 ≤ n, k ≤ 10 ^ 5 1 ≤ x ≤ 10 ^ 5<br> Alguém me ajuda?
enot [183]

Answer:

ask your teacher for help

Step-by-step explanation:

8 0
2 years ago
Follow the process of completing the square to solve 2x2 + 8x - 12 = 0.
Rom4ik [11]
Please:  Use "^" to denote exponentiation:  <span>2x^2 + 8x - 12 = 0

Reduce this by div. every term by 2:             </span><span>x^2 + 4x - 6 = 0

Here a=1, b=4 and c = -6.  Square half of b, obtaining (4/2)^2 = 4, and add, and then subtract, this 4 to x^2 + 4x - 6:

</span> x^2 + 4x +4  - 4 - 6 = 0.  Rewrite the square as (x+2)^2, obtaining new equation

(x+2)^2 = 10.  Take the sqrt of both sides:   x+2 = plus or minus sqrt(10).

Finally, solve for x:  x = -2 plus or minus sqrt(10).



8 0
3 years ago
HI PLEASE HELP ME WITH MY CALCULUS 1 HW? I AM REALLY STUCK. I need help with parts d,e,g.
asambeis [7]

(d) The particle moves in the positive direction when its velocity has a positive sign. You know the particle is at rest when t=0 and t=3, and because the velocity function is continuous, you need only check the sign of v(t) for values on the intervals (0, 3) and (3, 6).

We have, for instance v(1)\approx-0.91 and v(4)\approx0.91>0, which means the particle is moving the positive direction for 3, or the interval (3, 6).

(e) The total distance traveled is obtained by integrating the absolute value of the velocity function over the given interval:

\displaystyle\int_0^6|v(t)|\,\mathrm dt=\int_0^3-v(t)\,\mathrm dt+\int_3^6v(t)\,\mathrm dt

which follows from the definition of absolute value. In particular, if x is negative, then |x|=-x.

The total distance traveled is then 4 ft.

(g) Acceleration is the rate of change of velocity, so a(t) is the derivative of v(t):

a(t)=v'(t)=-\dfrac{\pi^2}9\cos\left(\dfrac{\pi t}3\right)

Compute the acceleration at t=4 seconds:

a(t)=\dfrac{\pi^2}{18}\dfrac{\rm ft}{\mathrm s^2}

(In case you need to know, for part (i), the particle is speeding up when the acceleration is positive. So this is done the same way as part (d).)

6 0
3 years ago
Find the solutions to x² = 12.
Novosadov [1.4K]

Answer:

x = \pm 2\sqrt3

Step-by-step explanation:

Hello!

Let's use the square root property to solve this question

Solve:

  • x^2 = 12
  • \sqrt{x^2} = \sqrt{12}
  • x = \pm\sqrt{12}
  • x = \pm \sqrt{4 * 3}
  • x = \pm 2\sqrt3

The answer is option A: x = ±2√3

3 0
2 years ago
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