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3 years ago

0.7 (3s+4) − 1.1s = 7.9

2 answers:

7 0

Simplify to 2.1s + 2.8 - 1.1s = 7.9. Simplify again to s = 5.1. If this is right could you possibly give me brainliest? Hope this helped.

irina1246 [14]3 years ago

4 0

Answer:

5.1

it is 5.1 my quiz with this had it on their and it said it was right

Step-by-step explanation:

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What’s the answer to this question? I’m confused as to how to do this.

choli [55]

Answer:

x = 20

Step-by-step explanation:

(3x + 50) = (6x - 10)

Subtract 6x on each side

3x + 50 - 6x = -10

Combine the x values together

-3x + 50 = -10

Subtract 50 on each side

-3x = -60

Divide each side by -3

x = 20

8 0

3 years ago

Binomial Expansion/Pascal's triangle. Please help with all of number 5.

Mandarinka [93]

$\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}$

The rows add up to $1,2,4,8,16$ , respectively. (Notice they're all powers of 2)

The sum of the numbers in row $n$ is $2^{n-1}$ .

The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When $n=1$ ,

$(1+x)^1=1+x=\dbinom10+\dbinom11x$

so the base case holds. Assume the claim holds for $n=k$ , so that

$(1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k$

Use this to show that it holds for $n=k+1$ .

$(1+x)^{k+1}=(1+x)(1+x)^k$
$(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)$
$(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}$

Notice that

$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}$

So you can write the expansion for $n=k+1$ as

$(1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}$

and since $\dbinom{k+1}0=\dbinom{k+1}{k+1}=1$ , you have

$(1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}$

and so the claim holds for $n=k+1$ , thus proving the claim overall that

$(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n$

Setting $x=1$ gives

$(1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n$

which agrees with the result obtained for part (c).

4 0

3 years ago

This math promblem!!!!

fiasKO [112]

1. You convert all the numbers into decimals.
a. For 8 1/9 you multiply 8x9 and add the numerator which in this case is one, so the equation would be 8x9=72 then 72+1= 73
b. For 81/10 I used a calculator for accuracy and I just divided 81 by 10 because the fraction line can also be used as a division sign. For this I got 8.1
2. Now I looked at all the numbers I had including the fractions I converted to decimals... 8.115, 8.55, 73, and 8.1
3. Lastly, I put the numbers in order from least to greatest: 8.1, 8.115, 8.55, and 73
4. In order to figure out which one is the smallest and largest, I just added zeros on the end of the numbers so they would all be the same: 8.1-->8.100, 8.115 I kept the same because it already had 3 decimal places, 8.55--> 8.550, and 73--> 73.000
5. Then i could tell which number was the largest by the decimal place numbers.
**Hope this was helpful... It's kind of hard to explain online but hopefully you have a better understanding of how to do it!**

6 0

3 years ago

Read 2 more answers

One eighth to a decimal

DochEvi [55]

Divide the numerator by the denominator and you will get 0.125

4 0

3 years ago

Read 2 more answers

Question: You can work 40 hours per pay period while attending school earning $12.50 an hour. There are two pay periods per mont

77julia77 [94]

1) Based on the information, the following taxation calculations can be made:

a) Gross Pay per pay period: $500

b) Taxable Income per pay period: $425

c) Since you pay 11% in federal taxes, the Federal tax withholding per pay period is <u>$46.75</u>.

2) The State tax withholding per pay period is <u>$20</u>.

3) The calculation of FICA, Medicare, and the Net Pay per pay period is as follows:

FICA withholding per pay period: $25.50

Medicare withholding per pay period: $6.16

Net Pay per pay period = $326.59

<h3>Data and Calculations:</h3>

Total hours worked per pay period = 40 hours

Rate per hour = $12.50

Pay periods per month = 2

Gross pay per pay period = $500 (40 x $12.50)

<h3>Deductions per pay period:</h3>

Health care = $25

401K deduction = $50

Total deductions per pay period = $75

Taxable income per pay period = $425 ($500 - $75)

Federal taxes = 11%

Federal tax withholding per pay period: $46.75 ($425 x 11%)

Annual taxable income = $10,200 ($425 x 24)

State withholding tax (annual) = $480 {$120 + ($10,200 - $5,000) x 5%}

State withholding tax per pay period = $20 ($480/24)

FICA withholding per pay period: $25.50 ($425 x 6.2%)

Medicare withholding per pay period: $6.16 ($425 x 1.45%)

<h3>Net Pay per pay period:</h3>

Gross Pay $500

Total deductions $75

Taxable income = $425

Federal Withholding = $46.75

State Withholding = $20

]FICA Withholding = $25.50

Medicare Withholding = $6.16

Net Pay per pay period = $326.59

Learn more about computing the net pay at brainly.com/question/15858747

#SPJ1

6 0

1 year ago