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Lilit [14]
3 years ago
14

George tells you that when variables are in the denominator, the equation becomes unsolvable. "There is a value for x that makes

the denominator zero, and you can't divide by zero," George explains. Using complete sentences, demonstrate to George how the equation is still solvable.
Mathematics
1 answer:
lyudmila [28]3 years ago
7 0
The equation can still be solved but it should be stated clearly at what value of would the equation be undefined. For instance, you have an expression,

1/x+2

When x is equal to -2, the denominator will be 0 resulting to an undefined solution. The value of x here can be any real number except -2.
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Math???????????.../?
Shalnov [3]
8.05 \cdot 10^8 - 9.16 \cdot 10^6 \\\\ =10^6(8.05 \cdot 10^2-9.16) \\\\ =10^6(805-9.16) \\\\ =10^6(795.84) \\\\ =795.84 \cdot 10^6 \\\\ = 7.96 \cdot 10^8\ kilograms
8 0
3 years ago
Read 2 more answers
Raise the following fractions to a higher term...
Pavlova-9 [17]

Answer:

1. B

2. B

Step-by-step explanation:

To find the missing term, find the scale factor between the fractions - the number which multiplies to one to make the other.

1. 7 x __ = 35

7x 5 = 35 therefore 5 is the scale factor. Multiply 15 x 5 = 75

2. 8 x ___ = 32

8 x 4 - 32 therefore the scale factor is 4. Multiply 1 x 4 = 4.

7 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
What is the pattern in this sequence? −3,1,5,9,13, ...
geniusboy [140]

Answer:

Pattern is: +4

6 0
3 years ago
Read 2 more answers
Please hellllpppp really appreciated
antiseptic1488 [7]
2/5 divides by 4/1 = 1/10
X=1/10
5 0
3 years ago
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