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quester [9]
3 years ago
13

How many parts 1/6s does it take to name the same amount as 1/3

Mathematics
2 answers:
Alecsey [184]3 years ago
7 0

Answer:

Two 1/6 pieces

Step-by-step explanation:

Two 1/6 pieces equals 1/3 of the apple

creativ13 [48]3 years ago
5 0
You need two 1/6s to get the same amount as 1/3.

So that will be 1/6 + 1/6 = 2/6
And 2/6 simplified is 1/3.
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The side of a square measures 3mn^2. What is the area of the square if m=4 and n= -2
Effectus [21]
Your question was wrong. I had to watch the photo.

-2²=4
4(4)=16
16(3)=48
48²=2304


When I read your text I though this:
4(-2)=-8
-8(3)=-24
-24²=576
576²=331 776

4 0
3 years ago
Help me please I really need help
Troyanec [42]

Answer: Find the area of all figures and then add it together

Step-by-step explanation:

5 0
3 years ago
Graph the function h(x) = -2x– 3.
Pani-rosa [81]

Answer:

look below

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Simplify (3x + 5) + (2x - 9) - (4x+3).<br><br><br> A.) X-7<br> B.) -X-7<br> C.) 9x1<br> D) 9x1
erica [24]

Answer:

x -7

Step-by-step explanation:

(3x + 5) + (2x - 9) - (4x+3)

3x + 5 + 2x - 9 -4x -3

x -7

8 0
3 years ago
Read 2 more answers
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
3 years ago
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