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MariettaO [177]

3 years ago

13

Elijah is selling coupon books for a school fundraiser. the coupon books sell for $11. the school gets $8 and $3 goes to elijah'

s homeroom. if elijah sold 24 coupon books, how much money did he collect? how much went to his homeroom? how much went to the school? help

1 answer:

Sladkaya [172]3 years ago

6 0

He collected $264.00, $72.00 went to his homeroom, and $192.00 to the school

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Evaluate x/y + 5 w^2, for x = 2/3 and y = 5/6, and w = 3

tresset_1 [31]

2/3 / 5/6 + 5 (3)^2

= 2/3 * 6/5 + 45

= 12/15 + 45

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5 0

4 years ago

(x +y)^5<br> Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=x,\:\:b=y$

$=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$

so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

$=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5$

$=1\cdot \:1\cdot \:x^5$

$=x^5$

also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5x^4y}{1}$

$=\frac{5x^4y}{1}$

$=5x^4y$

similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

$=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

Therefore,

$\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

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3 years ago

Which of these polynomials will make the following equation true<br><br>

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Poop I think good luck but it’s b x2-
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3 years ago

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5 0

3 years ago

The three consecutive terms

Studentka2010 [4]

Answer:

<h3>1</h3>

Step-by-step explanation:

The nth term of an exponential sequence is expressed as ar^n-1

The nth term of a linear sequence is expressed as Tn = a + (n-1)d

a is the first term

r is the common ratio

d is the common difference

n is the number of terms

Let the three consecutive terms of an exponential sequence be a/r, a and ar

second term of a linear sequence = a +d

third term of a linear sequence = a + 2d

sixth term of a linear sequence = a + 5d

Now if the three consecutive terms of an exponential sequence are the second third and sixth terms of a linear sequence, this is expressed as;

a/r = a + d ..... 1

a = a + 2d ..... 2

ar = a+ 5d .... 3

From 2: a = a + 2d

a-a= 2d

0 = 2d

d = 0/2

d = 0

Substitute d = 0 into equation 1:

From 1: a/r = a + d

a/r = a+0

a/r = a

Cross multiply

a = ar

a/a = r

1 = r

Rearrange

r = 1

<em>Hence the common ratio of the exponential sequence is 1</em>

5 0

3 years ago