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10 months ago

3х – 8y = -56 -3х – 56 = -8ysolve using system of equations

1 answer:

7 0

$\begin{gathered} 3x-8y=-56 \\ -3x+8y=56 \\ \text{those equations are equivalent, so we only need to solve one},\text{ } \\ 3x-8y=-56\rightarrow3x=-56+8y\rightarrow x=\frac{-56+8y}{3} \\ \text{let y=t} \\ S=\lbrace(\frac{-56+8t}{3},t)^{}\colon t\in R\}^{} \end{gathered}$

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1-What is the sum of the series? ∑j=152j Enter your answer in the box.

tangare [24]

Answer:

Please see the Step-by-step explanation for the answers

Step-by-step explanation:

∑ $\left \ {{5} \atop {j=1}} \right.$ 2j

The sum of series from j=1 to j=5 is:

∑ = 2(1) + 2(2) + 2(3) + 2(4) + 2(5)

= 2 + 4 + 6 + 8 + 10

∑ = 30

This question is not given clearly so i assume the following series that will give you an idea how to solve this:

∑ $\left \ {{4} \atop {k=1}} \right.$ 2k²

The sum of series from k=1 to j=4 is:

∑ = 2(1)² + 2(2)² + 2(3)² + 2(4)²

= 2(1) + 2(4) + 2(9) + 2(16)

= 2 + 8 + 18 + 32

∑ = 60

∑ $\left \ {{4} \atop {k=1}} \right.$ (2k)²

∑ = (2*1)² + (2*2)² + (2*3)² + (2*4)²

= (2)² + (4)² + (6)² + (8)²

= 4 + 16 + 36 + 64

∑ = 120

∑ $\left \ {{4} \atop {k=1}} \right.$ (2k)²- 4

∑ = (2*1)²-4 + (2*2)²-4 + (2*3)²-4 + (2*4)²-4

= (2)²-4 + (4)²-4 + (6)²-4 + (8)²-4

= (4-4) + (16-4) + (36-4) + (64-4)

= 0 + 12 + 32 + 60

∑ = 104

∑ $\left \ {{4} \atop {k=1}} \right.$ 2k²- 4

∑ = 2(1)²-4 + 2(2)²-4 + 2(3)²-4 + 2(4)²-4

= 2(1)-4 + 2(4)-4 + 2(9)-4 + 2(16)-4

= (2-4) + (8-4) + (18-4) + (32-4)

= -2 + 4 + 14 + 28

∑ = 44

∑ $\left \ {{6} \atop {k=3}} \right.$ (2k-10)

∑ = (2×3−10) + (2×4−10) + (2×5−10) + (2×6−10)

= (6-10) + (8-10) + (10-10) + (12-10)

= -4 + -2 + 0 + 2

∑ = -4

1+1/2+1/4+1/8+1/16+1/32+1/64

This is a geometric sequence where first term is 1 and the common ratio is 1/2 So

a = 1

This can be derived as

1/2/1 = 1/2 * 1 = 1/2

1/4/1/2 = 1/4 * 2/1 = 1/2

1/8/1/4 = 1/8 * 4/1 = 1/2

1/16/1/8 = 1/16 * 8/1 = 1/2

1/32/1/16 = 1/32 * 16/1 = 1/2

1/64/1/32 = 1/64 * 32/1 = 1/2

Hence the common ratio is r = 1/2

So n-th term is:

$ar^{n-1}$ = $1(\frac{1}{2})^{n-1}$

So the answer that represents the series in sigma notation is:

∑ $\left \ {{7} \atop {j=1}} \right.$ $(\frac{1}{2})^{j-1}$

−3+(−1)+1+3+5

This is an arithmetic sequence where the first term is -3 and the common difference is 2. So

a = 1

This can be derived as

-1 - (-3) = -1 + 3 = 2

1 - (-1) = 1 + 1 = 2

3 - 1 = 2

5 - 3 = 2

Hence the common difference d = 2

The nth term is:

a + (n - 1) d

= -3 + (n−1)2

= -3 + 2(n−1)

= -3 + 2n - 2

= 2n - 5

So the answer that represents the series in sigma notation is:

∑ $\left \ {{5} \atop {j=1}} \right.$ (2j−5)

6 0

3 years ago

What is the solution to this system of linear equations?

svetoff [14.1K]

It’s B
If y-x=6
Y +X =_10
then y= 6 + x, instead of y insert this no
6+ x+ x =-10
6 + 2x =-10 then collect like terms
2x =-10-6
2x=-16 then multiple both side by 1/2
X=-8
Y=6+x instead of x insert -8
Y=6-8
=-2

7 0

3 years ago

PLEASE HELP ASAP!!

balu736 [363]

Answer:219

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

If you want to be 95% confident of estimating the population mean to within a sampling error of plus or minus3 and the standard

Aleksandr [31]

Answer:

The sample size required is at least 171

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

What sample size is required?

A samle size of at least n is required, in which n is found when $M = 3, \sigma = 20$

$M = z*\frac{\sigma}{\sqrt{n}}$

$3 = 1.96*\frac{20}{\sqrt{n}}$

$3\sqrt{n} = 20*1.96$

$\sqrt{n} = \frac{20*1.96}{3}$

$(\sqrt{n})^{2} = (\frac{20*1.96}{3})^{2}$

$n = 170.7$

Rouding up

The sample size required is at least 171

5 0

3 years ago

PLEASE HELP!!! ILL MARK A HIGH POINT COUNT, GIVE BRAINLIEST, FIVE STARS, AND GIVE THANKS TO WHOEVER ANSWERS CORRECTLY!!!!

Westkost [7]

Each letter of the alphabet is worth two times as much as the one before it, implying that the value of each letter rises in mathematical progression. The formula for finding the nth term of an arithmetic progression would be used. I am written as

a + (n - 1)d = Tn

Where

The number of terms in the arithmetic sequence is represented by n.

The common difference of the terms in the arithmetic sequence is represented by d.

The first term of the arithmetic sequence is represented by a.

Tn stands for the nth word.

Based on the facts provided,

n = 26 characters1 Equals a

3 minus 1 equals 2 (difference between 2 letters)

Therefore,

1 + (26 - 1)2 = T26

51 = T26

The formula for calculating the sum of an arithmetic sequence's n terms

is as follows:

[2a + (n - 1)d] Sn = n/2

As a result, S26 is the sum of the first 26 terms.

S26 = 20/2[2 1 + (26 - 1)2] S26 = 20/2

[2 + 50] S26 =

676 = S26 = 13 52

3 0

1 year ago