Help me with this question

NEED TO SHOW ALL WORK) For each of the equations given below, use the triangle method to solve for the

Lelechka [254]

Answer:

1) $a=\dfrac{F}{m}$

2) $m=\dfrac{p}{v}$

Step-by-step explanation:

1) The given equation is

$F=ma$

We need to solve this for a.

In the given triangle place single variable i.e., F, on the top and variables in the product in the bottom (order of m and a does not matter) as shown in the below figure.

Using the triangle and the operations (× and ÷), we get

$F=ma$

$a=\dfrac{F}{m}$

$m=\dfrac{F}{a}$

Therefore, the required answer is $a=\dfrac{F}{m}$ .

2) The given equation is

$p=mv$

We need to solve this for m.

In the given triangle place single variable i.e., p, on the top and variables in the product in the bottom (order of m and v does not matter) as shown in the below figure.

Using the triangle and the operations (× and ÷), we get

$p=mv$

$m=\dfrac{p}{v}$

$v=\dfrac{p}{m}$

Therefore, the required answer is $m=\dfrac{p}{v}$ .

3 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.

7 0

2 years ago

Jack is using a ladder to hang lights up in his house.he places the ladder 5 feet from the base of his house and leans it so it

Yakvenalex [24]

You did not complete your question but I think this is what you are looking for. View the image.

5 0

2 years ago

I will give brainaliest if answer is correct.a baseball diamond is square.the distance from home plate to first base is 90 feet.

yan [13]

It would be 127 (usually a little over)

3 0

3 years ago

Write about a real-world situation that can be solved using a two-step equation. Then write the equation and solve the problem.

otez555 [7]

Hmm idk if this is what you need, but here we go: Amy is 5 years older than her brother, John. Her age is 19. How old is John?
x-John
x+5=19
x=19-5=14
John is 14.

4 0

3 years ago