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AysviL [449]
3 years ago
5

A gym membership at A costs $10/month plus a one-time membership fee of $15, and at B, the memberships costs $4/month plus a one

-time $40 fee. After about how many months will the gym memberships cost the same amount?
Mathematics
1 answer:
lora16 [44]3 years ago
5 0

Answer: it will take about 5 months for the cost to be the same

Step-by-step explanation:

Let x represent the number of months it will take for the gym memberships to cost the same amount.

A gym membership at A costs $10/month plus a one-time membership fee of $15. This means that the total cost for x months would be

10x + 15

At B, the memberships costs $4/month plus a one-time $40 fee. This means that the total cost for x months would be

4x + 40

For the costs to be the same,

10x + 15 = 4x + 40

10x - 4x = 40 - 15

6x = 25

x = 25/6 = 4.2

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0.42 - 4) = 0.24 + 7)
Ronch [10]

Answer:

im sorry if not but i think its 7.24

Step-by-step explanation:

8 0
2 years ago
Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.
Vera_Pavlovna [14]

Split up the integration interval into 4 subintervals:

\left[0,\dfrac\pi8\right],\left[\dfrac\pi8,\dfrac\pi4\right],\left[\dfrac\pi4,\dfrac{3\pi}8\right],\left[\dfrac{3\pi}8,\dfrac\pi2\right]

The left and right endpoints of the i-th subinterval, respectively, are

\ell_i=\dfrac{i-1}4\left(\dfrac\pi2-0\right)=\dfrac{(i-1)\pi}8

r_i=\dfrac i4\left(\dfrac\pi2-0\right)=\dfrac{i\pi}8

for 1\le i\le4, and the respective midpoints are

m_i=\dfrac{\ell_i+r_i}2=\dfrac{(2i-1)\pi}8

  • Trapezoidal rule

We approximate the (signed) area under the curve over each subinterval by

T_i=\dfrac{f(\ell_i)+f(r_i)}2(\ell_i-r_i)

so that

\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4T_i\approx\boxed{3.038078}

  • Midpoint rule

We approximate the area for each subinterval by

M_i=f(m_i)(\ell_i-r_i)

so that

\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4M_i\approx\boxed{2.981137}

  • Simpson's rule

We first interpolate the integrand over each subinterval by a quadratic polynomial p_i(x), where

p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}

so that

\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx

It so happens that the integral of p_i(x) reduces nicely to the form you're probably more familiar with,

S_i=\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{r_i-\ell_i}6(f(\ell_i)+4f(m_i)+f(r_i))

Then the integral is approximately

\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4S_i\approx\boxed{3.000117}

Compare these to the actual value of the integral, 3. I've included plots of the approximations below.

3 0
3 years ago
One day, a sprinter runs the 100 m dash 24 times, for a total distance of 24 00 m. how many kilometers did the sprinter run?
tigry1 [53]

The total distance that was covered by the sprinter in Kilometers is:

  • 2.4 Km

<h3>How much is the total distance covered?</h3>

To obtain the total distance covered by the sprinter, we need to first realize that 1 Km is equal to 1000 meters. That said, the question above requires us to change 2400 meters to kilometers.

Thus we will have:

2400 Meters * 1 km / 1000 meters

= 2.4 km

So, the total distance covered by the sprinter in kilometers is 2.4

Learn more about meter conversions here:

brainly.com/question/26259781

#SPJ4

4 0
1 year ago
WILL MARK BRAINLEIST!! Each figure is composed of large squares and small squares. The side length of the
julsineya [31]

9514 1404 393

Answer:

  A. x² -4

  B. 2x² +8

Step-by-step explanation:

The area of a square is the square of the side lengths. That means the area of each large square is x², since it has a side length of x.

__

A. The area of 4 small squares is subtracted from the area of the large square to find the shaded area:

  x² -4×1² = x² -4 . . . area of Figure A

__

B. The area of two small squares is added to the area of two large squares to find the shaded area:

  2x² +2×2² = 2x² +8 . . . area of Figure B

5 0
3 years ago
What is the distance between 15 miles east and 12 miles west?
inysia [295]

Answer:

27 miles.

Step-by-step explanation:

if you count from the 15 miles from the east to the 12 miles in the west, it would add up to 27. hope this helped

7 0
2 years ago
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