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3 years ago

9

Ms.sanders is buying three 112-oz bottles of laundry detergent for her family. Each bottle has the same price. The total cost of

the detergent , including $1.60 in sales tax, is $41.65 . To the nearest cent what price per ounce of the detergent before tax?

2 answers:

Zigmanuir [339]3 years ago

8 0

Let x represent the price of laundry detergent
3x+1.6=41.65
3x=40.05
x=$13.35 divided 112=$0.12. As a result, the detergent costs $0.12 per ounce. Hope it help!

Vaselesa [24]3 years ago

3 0

It would be 11 cents

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3 years ago

PLZ HELP ME, THIS IS URGENT! I’m completely lost... :(

Leno4ka [110]

Let’s solve this step by step. :)

First, let’s assign variables for the missing number: x and y.

so the 3 numbers would be x, y and 28

then we have 3 scenarios:
28 + x = 61
28 + y = 76
x + y = 81

Using the first equation, we can get x = 61-28 = 33

Using the 2nd equation, we can get y = 76-28 = 48

then, we can verify if our answer is right by substituting these values in the 3rd equation:

x + y = 81
33 + 48 = 81
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So the 3 numbers are 28, 33 and 48.

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3 years ago

Given the recursive formula for an arithmetic sequence find the common difference and the 52nd term

Mice21 [21]

Answer:

Common difference(d) $a_{52}$

(21) -10 -548

(22) -7 -323

(23) 10 547

(24) -100 -5118

Step-by-step explanation:

Let the common difference be denoted by 'd'.

Also the nth difference of an arithmetic sequence is given by: $a_{n}=a_{1}+(n-1)\times d$

(21)

We are given a recursive formula as:

$a_{n}=a_{n-1}-10$

The first term is given by:

$a_{1}=-38$

The common difference for an arithmetic sequence is given by:

$a_{n}-a_{n-1}$

Hence, here we have the common difference as:

$d=-10$

The nth term of an arithmetic sequence is given by:

$a_{n}=a_{1}+(n-1)\times d$

Here $a_{1}=-38$ and $d=-10$ .

Hence, $a_{52}=-38+(52-1)\times (-10)$

Hence, $a_{52}=-548$

(22)

$a_{n}=a_{n-1}-7$

$a_{1}=34$

The common difference for an arithmetic sequence is given by:

$a_{n}-a_{n-1}$

Hence, here we have the common difference as:

$d=-7$

Here $a_{1}=34$ and $d=-7$ .

Hence, $a_{52}=34+(52-1)\times (-7)$

Hence, $a_{52}=-323$

(23)

$a_{n}=a_{n-1}+10$

$a_{1}=37$

The common difference for an arithmetic sequence is given by:

$a_{n}-a_{n-1}$

Hence, here we have the common difference as:

$d=10$

Here $a_{1}=37$ and $d=10$ .

Hence, $a_{52}=37+(52-1)\times (10)$

Hence, $a_{52}=547$

(24)

$a_{n}=a_{n-1}-100$

$a_{1}=-18$

The common difference for an arithmetic sequence is given by:

$a_{n}-a_{n-1}$

Hence, here we have the common difference as:

$d=-100$

Here $a_{1}=-18$ and $d=-100$ .

Hence, $a_{52}=-18+(52-1)\times (-100)$

Hence, $a_{52}=-5118$

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4 years ago