Are 4/19 and -4 3/4 multiplicative inverse

Solve by elimination: 3y-4x=29 2y+5=4

Vaselesa [24]

The solution is x = -61/8 and y = -1/2.

Step-by-step explanation:

The given equations are -4x + 3y = 29 and 2y + 5 = 4

Step 1 : There is no x term in the second equation.

Step 2 : Solve for y, 2y = 4 -5

y = -1/2

Step 3 : Substitute y = -1/2 in first equation and solve for x.

Step 4 : -4x + 3(-1/2) = 29

-4x -(3/2) = 29

-8x -3 = 29 $\times$ 2

-8x = 58 + 3

x = -61/8

5 0

2 years ago

Hi I need help with this question it says An auditorium has 25 rows with 45 seats in each row. How many seats are there in all.

pantera1 [17]

there are 1,125 seats total.

4 0

3 years ago

Read 2 more answers

Please solve no links i will mark brailest

Lemur [1.5K]

Answer:

oki! Mizuki here to help! 12 will be your answer!

Step-by-step explanation:

Soo... The ratio of blue to green is 4:3 right?

and there are 21 balloons!

4 + 3 is 7! So we divide 21 by 7 and get 3!

Then we multiply 3 by the number of blue balloons(aka 4) and get 12!

3 0

3 years ago

7: Which of the following numbers is smallest?

iogann1982 [59]

Answer:

d) 0.712

Step-by-step explanation:

Your question isn't clear, so my answer may be wromg

6 0

2 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

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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.

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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.

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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.

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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.

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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.

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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