Find the nth term<br><br> 1, 5, 9. 13, 17

Ken's fitness member ship cost $100 to join plus a $25 monthly fee. Ken also used a coupon to get 10% off the monthly fee. What

GREYUIT [131]

The equation that models the total cost is y = 100 + 22.5m

Step-by-step explanation:

Fitness member ship cost = $100

Monthly fee = $25

Discount = 10% on monthly fee

To calculate new monthly fee after discount,

25 - (25*10%) --> 25 - 2.5 --> $22.5

Number of months = m

Fee in m months = 22.5*m = 22.5m

<u>Equation:</u>

(c) is constant which is the membership fee

(x) is the monthly fee $22.5

(m) is the number of months

(y) is the total cost

So,

y = mx + c

y = m(22.5) + 100

y = 100 + 22.5m

Therefore, the equation that models the total cost is y = 100 + 22.5m

Keyword: Equation

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5 0

3 years ago

Sophie pays m dollars to rent a car for the weekend on the drive home she buys $45 worth of gasoline her friend melody pays for

Hoochie [10]

Complete Question

Sophie pays m dollars to rent a car for the weekend. On the drive home, she buys $45 worth of gasoline. Her friend Melody pays for one-half the cost of the car rental only. Write an expression you could use to determine how much Sophie spent.

Answer:

1/2m + 45

Step-by-step explanation:

We are told in the question that:

On the drive home, she buys $45 worth of gasoline.

Her friend Melody pays for one-half the cost of the car rental only.

Sophie pays m dollars to rent a car for the weekend.

The amount sophie pays for car rental = m - 1/2m

= 1/2m

Therefore, the expression used to determine how much sophie spent is written as:

1/2 × $m + $45

1/2m + 45

4 0

3 years ago

Use the four-step definition of the derivative to find f'(x) if f(x) = −4x^3 −1.

guapka [62]

$\stackrel{de finition \textit{ of a derivative as a limit}}{\lim\limits_{h\to 0}~\cfrac{f(x+h)-f(x)}{h}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{[-4(x+h)^3-1]~~ - ~~[-4x^3-1]}{h} \\\\\\ \cfrac{[-4(x^3+3x^2h+3xh^2+h^3)-1]~~ ~~+4x^3+1}{h} \\\\\\ \cfrac{[-4x^3-12x^2h-12xh^2-4h^3-1]~~ ~~+4x^3+1}{h} \\\\\\ \cfrac{-4x^3-12x^2h-12xh^2-4h^3-1+4x^3+1}{h}\implies \cfrac{-12x^2h-12xh^2-4h^3}{h}$

$\cfrac{h(-12x^2-12xh-4h^2)}{h}\implies -12x^2-12xh-4h^2 \\\\\\ \lim\limits_{h\to 0}~-12x^2-12xh-4h^2\implies \lim\limits_{h\to 0}~-12x^2-12x(0)-4(0)^2 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \lim\limits_{h\to 0}~-12x^2~\hfill$

7 0

2 years ago

Please hurry no clown answers and please do the whole thing

mr Goodwill [35]

Answer:

100-40=m Answer:60

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

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

Find the nth term 1, 5, 9. 13, 17