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3 years ago

Please help it’s due

Mathematics

1 answer:

aleksandr82 [10.1K]3 years ago

5 0

1 = 70

2 = 20

3 = 70

4 = 20

5 =110

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What is the maturity value of $1 at a compound interest rate of 9 percent over 4 periods

Vaselesa [24]

Your answer is 1. Hoped this has helped

5 0

3 years ago

A rectangular driveway has an area of 880 square feet and a perimeter of 128 feet. What are the dimensions of the driveway?

mamaluj [8]

Answer:

Dimensions are 20 X 44 feet

Step-by-step explanation:

First we need to build two equations:

For area X times Y = 880

For Perimeter 2x + 2y = 128

Now we need to solve one so we can substitute it into the other:

2x + 2y = 128 becomes

2x = 128 - 2y and then

x = 64 - y We will substitute this into the are equation

(64 - y)y = 880

64y - $y^{2}$ = 880

- $y^{2} + 64y -880 = 0$

factor out the negative

-( $y^{2} - 64y +880 = 0$

Factor:

-(y - 20)(y - 44) = 0

y-20 = 0 and y - 44 = 0

y = 20 and y = 44

Plugging these in two both of our formulas works

20 X 44 = 880 for area

2(20) + 2 (44) = 128 for perimeter

8 0

3 years ago

Show work explain with formulas.

Sedaia [141]

Answer:

$\large\boxed{1.\ a_{44}=481,\ \sum\limits_{n=1}^\infty(11n-3),\ \text{the sum not exist}}$

$\large\boxed{3.\ \sum\limits_{n=1}^{47}(9n+16)=10,904}$

$\large\boxed{4.\ \sum\limits_{n=1}^\infty\bigg(6(2^n)\bigg),\ \text{the sum not exist}}$

Step-by-step explanation:

$1.\\8+19+30+41+...\\\\19-8=11\\30-19=11\\41-30=11\\\\\text{It's an arithmetic series.}\ a_1=8,\ d=11.\\\\\text{The formula for the n-th term of an arithmetic sequence:}\\\\a_n=a_1+(n-1)d\\\\\text{Substitute:}\\\\a_n=8+(n-1)(11)\qquad\text{use the distributive property}\ a(b-c)=ab-ac\\a_n=8+11n-11\\\boxed{a_n=11n-3}\\\\\text{Calculate the 44th term. Put n = 44 to the formula:}\\\\a_{44}=(11)(44)-3=484-3=481$

$\text{The summation notation:}\\\\\sum\limits_{n=1}^\infty(11n-3)\\\\\text{The sum not exist, because}\ d>1,\ \text{therefore}\ 11n-3\rightarrow\infty.$

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$3.\\25+34+43+52+...+436\\\\34-25=9\\43-34=9\\52-43=9\\\\\text{It's an arithmetic series.}\ a_1=25,\ d=9.\\\\\text{The formula for the n-th term of an arithmetic sequence:}\\\\a_n=a_1+(n-1)d\\\\\text{Substitute:}\\\\a_n=25+(n-1)(9)\\a_n=25+9n-9\\a_n=9n+16\\\\\text{Calculate which the term of arithmetic series is the number 439.}\\\text{Put}\ a_n=439:\\\\9n+16=439\qquad\text{subtract 16 from both sides}\\9n=423\qquad\text{divide both sides by 9}\\n=47$

$\text{Therefore we have the series in summation notation:}\\\\\sum\limits_{n=1}^{47}(9n+16)\\\\\text{For calculation of the sum we use the formula of a sum of terms}\\\text{of an arithmetic sequence:}\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\text{Substitute}\ n=47,\ a_1=25,\ a_{47}=439:\\\\S_{47}=\dfrac{25+439}{2}\cdot47=\dfrac{464}{2}\cdot47=232\cdot47=10904\\\\\sum\limits_{n=1}^{47}(9n+16)=10,904$

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$4.\\12+24+48+...\\\\24:12=2\\48:24=2\\\\\text{It's\ a\ geometric series}\ a_1=12,\ r=2.\\\\\text{The formula for the n-th term of a geometic sequence:}\\\\a_n=a_1r^{n-1}\\\\\text{Substitute:}\\\\a_n=(12)(2^{n-1})\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\a_n=(12)\left(\dfrac{2^n}{2^1}\right)\\\\a_n=(6)(2^n)\\\\\text{The summation notation:}\\\\\sum\limits_{n=1}^\infty\bigg(6(2^n)\bigg)\\\\\text{The sum not exist, because}\ r>1,\ \text{therefore}\ 6(2^n)\to\infty$