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4 years ago

11

If c1 and c2 are real numbers, give the relationship between them that will allow you to find solutions for the equation xn + c1

= c2, regardless of the value of n.

1 answer:

quester [9]4 years ago

4 0

If xn+c1=c2 then xn+c2=c1

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At a bargain store, Spongebob bought 3 items that each cost the same amount. Patrick bought 5 items that each cost the same amou

STALIN [3.7K]

Answer:

Cost of each item bought by Spongebob $=x=\$3.75$

Cost of items bought by Spongebob = $3x=3(3.75)=\$11.25$

Cost of each item bought by Patrick $=x-1.50=3.75-1.50=\$2.25$

Cost of items bought by Patrick = $5(x-1.50)=5(3.75-1.50)=\$11.25$

Step-by-step explanation:

Let $x$ denotes cost of each item bought by Spongebob.

Number of items bought by Spongebob = 3

So,

Cost of items bought by Spongebob = $3x$

Cost of each item bought by Patrick = $x-1.50$

Number of items bought by Patrick = 5

Cost of items bought by Patrick = $5(x-1.50)$

Both Spongebob and Patrick paid the same amount of money.

So,

$3x=5(x-1.50)\\3x=5x-7.5\\5x-3x=7.5\\2x=7.5\\x=\frac{7.5}{2} \\x=3.75$

So,

Cost of each item bought by Spongebob $=x=\$3.75$

Cost of items bought by Spongebob = $3x=3(3.75)=\$11.25$

Cost of each item bought by Patrick $=x-1.50=3.75-1.50=\$2.25$

Cost of items bought by Patrick = $5(x-1.50)=5(3.75-1.50)=\$11.25$

6 0

3 years ago

A gaming system is marked down from 290 to 188. What is the percentage of the discount?

yuradex [85]

Let P = percent of discount

(290-188)/290 = P/100

102/290 = P/100

290P = (102)(100)

290P = 10,200

Solve for P to find your answer.

5 0

4 years ago

MAURICE put 130 training cards and protector sheets he feels seven sheets and puts the remaining four cards in an eighth sheet e

n200080 [17]

18 goes into 7 filled sheets. 4 on a single sheet.

4 0

3 years ago

In how many distinct ways can the letters of the word mathematics be arranged? (first, does the order matter?)

TEA [102]

Our current list has 11!/2!11!/2! arrangements which we must divide into equivalence classes just as before, only this time the classes contain arrangements where only the two As are arranged, following this logic requires us to divide by arrangement of the 2 As giving (11!/2!)/2!=11!/(2!2)(11!/2!)/2!=11!/(2!2).

Repeating the process one last time for equivalence classes for arrangements of only T's leads us to divide the list once again by 2

4 0

3 years ago

The picture up there need this for today!!

tatyana61 [14]

Answer:

I believe its 30 different ways.

Step-by-step explanation:

Each different topping can be added times. so 5 times 6 is 30.

3 0

3 years ago