Ashli divided 25 jolly ranchers into 5 small piles. She gave away 4/5 of one pile and ate 2 pieces. How much piles were left?

Let f(x)=4x-1 and g(x)=2x^2+3. Perform each function operations and then find the domain.

Triss [41]

F(x) = 4x - 1
g(x) = 2x² + 3

1. (f + g)(x) = (4x - 1) + (2x² + 3)
(f + g)(x) = 2x² + 4x + (-1 + 3)
(f + g)(x) = 2x² + 4x + 2
Domain: {x| -∞ < x < ∞}, (-∞, ∞)

2. (f - g)(x) = (4x + 1) - (2x² + 3)
  (f - g)(x) = 4x + 1 - 2x² - 3
(f - g)(x) = -2x² + 4x + 1 - 3
(f - g)(x) = -2x² + 4x - 2
Domain: {x|-∞ < x < ∞}, (-∞, ∞)
3. (g - f)(x) = (2x² + 3) - (4x - 1)
(g - f)(x) = 2x² + 3 - 4x + 1
(g - f)(x) = 2x² - 4x + 3 + 1
(g - f)(x) = 2x² - 4x + 4
Domain: {x| -∞ < x < ∞}, (-∞, ∞)

4. (f · g)(x) = (4x + 1)(2x² + 3)
(f · g)(x) = 4x(2x² + 3) + 1(2x² + 3)
(f · g)(x) = 4x(2x²) + 4x(3) + 1(2x²) + 1(3)
(f · g)(x) = 8x³ + 12x + 2x² + 3
(f · g)(x) = 8x³ + 2x² + 12x + 3
Domain: {x| -∞ < x < ∞}, (-∞, ∞)

5. $(\frac{f}{g})(x) = \frac{4x - 1}{2x^{2} + 3}$
Domain: 2x² + 3 ≠ 0
- 3 - 3
2x² ≠ 0
2 2
  x² ≠ 0
x ≠ 0
(-∞, 0) ∨ (0, ∞)

6. $(\frac{g}{f})(x) = \frac{2x^{2} + 3}{4x - 1}$
Domain: 4x - 1 ≠ 0
+ 1 + 1
4x ≠ 0
4   4
x ≠ 0
  (-∞, 0) ∨ (0, ∞)

6 0

3 years ago

Solve the following equation:

Rama09 [41]

Complete the square.

$z^4 + z^2 - i\sqrt 3 = \left(z^2 + \dfrac12\right)^2 - \dfrac14 - i\sqrt3 = 0$

$\left(z^2 + \dfrac12\right)^2 = \dfrac{1 + 4\sqrt3\,i}4$

Use de Moivre's theorem to compute the square roots of the right side.

$w = \dfrac{1 + 4\sqrt3\,i}4 = \dfrac74 \exp\left(i \tan^{-1}(4\sqrt3)\right)$

$\implies w^{1/2} = \pm \dfrac{\sqrt7}2 \exp\left(\dfrac i2 \tan^{-1}(4\sqrt3)\right) = \pm \dfrac{2+\sqrt3\,i}2$

Now, taking square roots on both sides, we have

$z^2 + \dfrac12 = \pm w^{1/2}$

$z^2 = \dfrac{1+\sqrt3\,i}2 \text{ or } z^2 = -\dfrac{3+\sqrt3\,i}2$

Use de Moivre's theorem again to take square roots on both sides.

$w_1 = \dfrac{1+\sqrt3\,i}2 = \exp\left(i\dfrac\pi3\right)$

$\implies z = {w_1}^{1/2} = \pm \exp\left(i\dfrac\pi6\right) = \boxed{\pm \dfrac{\sqrt3 + i}2}$

$w_2 = -\dfrac{3+\sqrt3\,i}2 = \sqrt3 \, \exp\left(-i \dfrac{5\pi}6\right)$

$\implies z = {w_2}^{1/2} = \boxed{\pm \sqrt[4]{3} \, \exp\left(-i\dfrac{5\pi}{12}\right)}$

3 0

2 years ago

. Using the Binomial Theorem explicitly, give the 15th term in the expansion of (-2x + 1)^19

timofeeve [1]

Let's rewrite the binomial as:
$(1 - 2x)^{19}$

$\text{Binomial expansion:} (1 + x)^{n} = \sum_{r = 0}^n\left(\begin{array}{ccc}n\\r\end{array}\right) (x)^{r}$

Using the binomial expansion, we get:
$\text{Binomial expansion: } (1 - 2x)^{19} = \sum_{r = 0}^{19}\left(\begin{array}{ccc}19\\r\end{array}\right) (-2x)^{r}$

For the 15th term, we want the term where r is equal to 14, because of the fact that the first term starts when r = 0. Thus, for the 15th term, we need to include the 0th or the first term of the binomial expansion.

Thus, the fifteenth term is:
$\text{Binomial expansion (15th term):} \left(\begin{array}{ccc}19\\14\end{array}\right) (-2x)^{14}$

3 0

4 years ago

Please help! I need to answer this last question, and it was due last night!! Will chose Brainliest!! Just find the missing meas

svp [43]

If it’s an angle I think 90% but if not maybe 24 all around

4 0

3 years ago

What's the best first step in solving-4 + 2/5 > 5/10

Artyom0805 [142]

Answer:

B

Step-by-step explanation:

There are 2 operations happening on the left side to x.

1. x is being multiplied by -4.

2. 2/5 is being added to -4x.

If you had a number for x, and you wanted to evaluate the left side, first, you would multiply the value of x by -4. Then you would add 2/5.

To solve an equation or an inequality, you do the opposite operation in the opposite order.

Since the second operation was add 2/5, to solve, you start by doing the opposite of the second operation. That means you subtract 2/5 from both sides.

Answer: B

3 0

3 years ago