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3 years ago

Which one would it be?

Mathematics

1 answer:

Leni [432]3 years ago

6 0

C is the answer.
in the original function, when x=0, y=-1
in the new function, to make y=-1, x+3=0, x=-3. From the original 0 to -3 is a shift of 3 units to the left.

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A pole that is 2.6 m tall casts a shadow that is 1.38 m long. At the same time, a nearby building casts a shadow that is 40.75 m

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Step-by-step explanation:

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PREALGEBRA

Melanie L. asked • 03/08/16

a pole that is 3.3m tall casts a shadow that is 1.78m long. at the same time, a nearby building casts a shoadow that is 49.75 m long. how tall is the building?

a pole that is 3.3m tall casts a shadow that is 1.78m long. at the same time, a nearby building casts a shoadow that is 49.75 m long. how tall is the building? Round to the nearest meter.

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Jane C. answered • 03/08/16

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Hi Melanie!

This problem involves proportion.

Let x be the unknown, which is the height of the building.

height of pole = height of the building

height of the shadow height of the shadow

3.3 m = x

1.78 m 49. 75 m

Solve for x.

(3.3 m) (49.75 m) = x

1.78 m

92.2331460674 = x

Rounded to the nearest meter, the building is 92 meters high.

Hope that helps.

The fraction bars are not coming out (Underline), I don't know why... but yeah, that is the answer.

8 0

3 years ago

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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3 years ago

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soldi70 [24.7K]

Answer:

Jaime's. Interval not centered around the point estimate.

Step-by-step explanation:

When constructing a confidence interval based on a point estimate, the obtained point estimate must be the central value of the interval.

For Jaime's interval

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