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V125BC [204]
2 years ago
7

The shorter leg of a right triangle is 18 meters. The hypotenuse is 6 meters longer than the longer leg. Find the length of the

longer leg.
Mathematics
1 answer:
mote1985 [20]2 years ago
6 0

Answer:

24 meters

Step-by-step explanation:

Use the pythagorean theorem

18² + x² = (x + 6)²

Expand

324 + x² = x² + 12x + 36

Subtract x² from both sides

324 = 12x + 36

Subtract 36 from both sides

288 = 12x

Divide both sides by 12

24 = x

24 meters

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Calculus hw, need help asap with steps.
nikdorinn [45]

Answers are in bold

S1 = 1

S2 = 0.5

S3 = 0.6667

S4 = 0.625

S5 = 0.6333

=========================================================

Explanation:

Let f(n) = \frac{(-1)^{n+1}}{n!}

The summation given to us represents the following

\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!}=\sum_{n=1}^{\infty} f(n)\\\\\\\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!}=f(1) + f(2)+f(3)+\ldots\\\\

There are infinitely many terms to be added.

-------------------

The partial sums only care about adding a finite amount of terms.

The partial sum S_1 is the sum of the first term and nothing else. Technically it's not really a sum because it doesn't have any other thing to add to. So we simply say S_1 = f(1) = 1

I'm skipping the steps to compute f(1) since you already have done so.

-------------------

The second partial sum is when things get a bit more interesting.

We add the first two terms.

S_2 = f(1)+f(2)\\\\S_2 = 1+(-\frac{1}{2})\\\\S_2 = \frac{1}{2}\\\\S_2 = 0.5\\\\\\

The scratch work for computing f(2) is shown in the diagram below.

-------------------

We do the same type of steps for the third partial sum.

S_3 = f(1)+f(2)+f(3)\\\\S_3 = 1+(-\frac{1}{2})+\frac{1}{6}\\\\S_3 = \frac{2}{3}\\\\S_3 \approx 0.6667\\\\\\

The scratch work for computing f(3) is shown in the diagram below.

-------------------

Now add the first four terms to get the fourth partial sum.

S_4 = f(1)+f(2)+f(3)+f(4)\\\\S_4 = 1+(-\frac{1}{2})+\frac{1}{6}-\frac{1}{24}\\\\S_4 = \frac{5}{8}\\\\S_4 \approx 0.625\\\\\\

As before, the scratch work for f(4) is shown below.

I'm sure you can notice by now, but the partial sums are recursive. Each new partial sum builds upon what is already added up so far.

This means something like S_3 = S_2 + f(3) and S_4 = S_3 + f(4)

In general, S_{n+1} = S_{n} + f(n+1) so you don't have to add up all the first n terms. Simply add the last term to the previous partial sum.

-------------------

Let's use that recursive trick to find S_5

S_5 = [f(1)+f(2)+f(3)+f(4)]+f(5)\\\\S_5 = S_4 + f(5)\\\\S_5 = \frac{5}{8} + \frac{1}{120}\\\\S_5 = \frac{19}{30}\\\\S_5 \approx 0.6333

The scratch work for f(5) is shown below.

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2 years ago
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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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The first quadrant is the upper right hand corner of the graph where both y and y are positive

the second quadrant in the upper left hand corner includes negative values of x and positive values of y

the third quadrant the lower he as corner includes negative values of both x and y

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SAT Math Question
mariarad [96]

I'm tentatively changing my answer to say this kind of relies on practical knowledge of how stores tend to operate. If 20 coupons are given out, the store has sold all 500 shirts, arguably at a loss to the retailer. They have to have more shirts in stock to be sold at full price because, well, that's how they make money. It's more likely that such a store would carry more than just 500 shirts at the start of each day, so A is (probably) wrong.

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