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Leno4ka [110]
3 years ago
8

What is the number sequence 3 4 6 9 13 18 24?

Mathematics
2 answers:
lawyer [7]3 years ago
7 0
Add 1, then add 2, then add 3, then 4, then  5, then 6. 
artcher [175]3 years ago
7 0
3+1=4
4+2=6
6+3=9
9+4=13
13+5=18
18+6=24
24+7=31
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There are 140 members in a community sports program. Sixty-five percent of these members play soccer.
defon
Well, we have to determinate the 65% of 140, which are 91 members play soccer. 

The total members are 140, if 91 play soccer, 49 members do not play soccer. 

The answer is 49 members.
3 0
3 years ago
20 divided by 4 - 5+12
likoan [24]
The answer is 12. you first divide 20 by 4 and get 5. then you subtract 5 by 5 and get 0. then you add 12 plus 0 and get 12

Hope I Helped :D




7 0
3 years ago
Read 2 more answers
Raul has been offered three different jobs and must choose which one to accept. One of the factors he is considering is the pay
chubhunter [2.5K]

Answer:

As the pay rate for Job C is the highest, so he must accept the offer from Job C.

Step-by-step explanation:

Given that Raul would work 50 weeks of the year,

40 hours each week,

5 days each week,

and he would receive a 1-hour lunch break each day.

Pay rate for Job A = $35,000 / year

  • For Job B:

Pay rate = $15 / hour

He will work 40 hours each week.

As he works 5 days a week, so in 1 day he will work 40/5 = 8 hours.

He would receive a 1-hour break each day, for which he will not be paid, so he will receive a payment for 8-1=7 hours.

So, payment for 1 day = $15 x 7 = $105

As he works 5 days a week, so payment for 1 week = $105 x 5 = $525

As he works 50 weeks a year, so payment for 1 year = $525 x 50 = $26250.

So, the pay rate by Job B = $26250 /year

  • For Job C:

Pay rate = $750 / week

As he works 50 weeks a year, so payment for 1 year = $750 x 50 = $37500.

So, the pay rate by Job C = $37500 /year

As the pay rate for Job C is the highest, so he must accept the offer from Job C.

8 0
3 years ago
Genber wiil evaluate an 8th degree polynomial in x at x=10 using the remainder theorem and synthetic division.how many coefficie
lisabon 2012 [21]
An 8th-degree polynomial needs 9 terms that involve
 x⁸, x⁷, ..., x¹, and x⁰.

x=10 implies that (x-10) is a factor of the polynomial according to the Remainder theorem.

Let the polynomial be of the form
f(x) = a₁x⁸ + a₂x⁷ + a₃x⁶ +a₄x⁵ + a₅x⁴ + a₆x³ + a₇x² + a₈x + a₉

The first few lines of the synthetic division are

10 | a₁  a₂  a₃  a₄  a₅  a₆  a₇  a₈  a₉             ( the first row has 9  coefficients)

    -----------------------------------------
      a₁

Answer:
The first row has 9  coefficients.
8 0
3 years ago
21. In 4 + In(4x - 15) = In(5x + 19)
Alexxx [7]

Answer:

x=-30;\quad \:I\ne \:0

Step-by-step explanation:

In\cdot \:4+In\left(4x-15\right)=In\left(5x+19\right)

\:4+In\left(4x-15\right):\quad -11nI+4nxI\\In\cdot \:4+In\left(4x-15\right)\\=4nI+nI\left(4x-15\right)\\\\\:In\left(4x-15\right):\quad 4nxI-15nI\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b-c\right)=ab-ac\\a=In,\:b=4x,\:c=15\\=In\cdot \:4x-In\cdot \:15\\=4nxI-15nI\\=In\cdot \:4+4nxI-15nI\\\mathrm{Simplify}\:In\cdot \:4+4nxI-15nI:\quad -11nI+4nxI\\In\cdot \:4+4nxI-15nI\\\mathrm{Group\:like\:terms}\\=4nI-15nI+4nxI\\\mathrm{Add\:similar\:elements:}\:4nI-15nI=-11nI\\=-11nI+4nxI\\

\mathrm{Expand\:}In\left(5x+19\right):\quad 5nxI+19nI\\In\left(5x+19\right)\\=nI\left(5x+19\right)\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac\\a=In,\:b=5x,\:c=19\\=In\cdot \:5x+In\cdot \:19\\=5nxI+19nI\\\\-11nI+4nxI=5nxI+19nI\\\\\mathrm{Add\:}11nI\mathrm{\:to\:both\:sides}\\-11nI+4nxI+11nI=5nxI+19nI+11nI\\Simplify\\4nxI=5nxI+30nI\\\mathrm{Subtract\:}5nxI\mathrm{\:from\:both\:sides}\\4nxI-5nxI=5nxI+30nI-5nxI\\\mathrm{Simplify}\\-nxI=30nI\\

\mathrm{Divide\:both\:sides\:by\:}-nI;\quad \:I\ne \:0\\\frac{-nxI}{-nI}=\frac{30nI}{-nI};\quad \:I\ne \:0\\\mathrm{Simplify}\\\frac{-nxI}{-nI}=\frac{30nI}{-nI}\\\mathrm{Simplify\:}\frac{-nxI}{-nI}:\quad x\\\frac{-nxI}{-nI}\\\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{-b}=\frac{a}{b}\\=\frac{nxI}{nI}\\\mathrm{Cancel\:the\:common\:factor:}\:n\\=\frac{xI}{I}\\\mathrm{Cancel\:the\:common\:factor:}\:I\\=x\\\mathrm{Simplify\:}\frac{30nI}{-nI}:\quad -30\\\mathrm{Apply\:the\:fraction\:rule}:

\quad \frac{a}{-b}=-\frac{a}{b}\\\mathrm{Cancel\:the\:common\:factor:}\:n\\=\frac{30I}{I}\\\mathrm{Cancel\:the\:common\:factor:}\:I\\=-30\\x=-30;\quad \:I\ne \:0

3 0
3 years ago
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