Answer:
<u>D. He recorded his paycheck amount for April 23rd incorrectly. Yes, this was Adam's mistake. The correct amount should be 341.60 and not 338.45.</u>
Step-by-step explanation:
1. Let's review the information given to us to help Adam to determine where his error is.
A. He completely forgot to include the clothes he bought from Bargains RUS. No, he didn't. It was recorded properly, including the sales tax amount.
B. He made an arithmetic error in the balance column. No he didn't, the balance was calculated correctly after each transaction.
C. He recorded one of his withdrawals in the deposit column. No, he didn't. All of the withdrawals are in the right column.
<u>D. He recorded his paycheck amount for April 23rd incorrectly. Yes, this was Adam's mistake. The correct amount should be 341.60 and not 338.45.</u>
The answer is: " 91 " .
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→ " B = 91 " .
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Explanation:
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Given:
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" A + B = 180 " ;
"A = -2x + 115 " ; ↔ A = 115 − 2x ;
"B = - 6x + 169 " ; ↔ B = 169 − 6x ;
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METHOD 1)
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Solve for "x" ; and then plug the solved value for "x" into the expression given for "B" ; to solve for "B"
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(115 − 2x) + (169 − 6x) =
115 − 2x + 169 − 6x = ?
→ Combine the "like terms" ; as follows:
+ 115 + 169 = + 284 ;
− 2x − 6x = − 8x ;
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And rewrite as:
" − 8x + 284 " ;
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→ " - 8x + 284 = 180 " ;
Subtract: "284" from each side of the equation:
→ " - 8x + 284 − 284 = 180 − 284 " ;
to get:
→ " -8x = -104 ;
Divide EACH SIDE of the equation by "-8 " ;
to isolate "x" on one side of the equation; & to solve for "x" ;
→ -8x / -8 = -104/-8 ;
→ x = 13
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Now, to find the value of "B" :
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"B = - 6x + 169 " ; ↔ B = 169 − 6x ;
↔ B = 169 − 6x ;
= 169 − 6(13) ; ===========> Plug in our "solved value, "13", for "x" ;
= 169 − (78) ;
= 91 ;
B = " 91 " .
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The answer is: " 91 " .
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→ " B = 91 " .
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Now; let us check our answer:
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→ A + B = 180 ;
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Plug in our "solved answer" ; which is "91", for "B" ; as follows:
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→ A + 91 = ? 180? ;
↔ A = ? 180 − 91 ? ;
→ A = ? -89 ? Yes!
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→ " A = -2x + 115 " ; ↔ A = 115 − 2x ;
Plug in our solved value for "x"; which is: "13" ;
" A = 115 − 2x " ;
→ A = ? 115 − 2(13) ? ;
→ A = ? 115 − (26) ? ;
→ A = ? 29 ? Yes!
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METHOD 2)
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Given:
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" A + B = 180 " ;
"A = -2x + 115 " ; ↔ A = 115 − 2x ;
"B = - 6x + 169 " ; ↔ B = 169 − 6x ;
→ Solve for the value of "B" :
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A + B = 180 ;
→ B = 180 − A ;
→ B = 180 − (115 − 2x) ;
→ B = 180 − 1(115 − 2x) ; ==========> {Note the "implied value of "1" } ;
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Note the "distributive property" of multiplication:__________________________________________________ a(b + c) = ab + ac ; <u><em>AND</em></u>:
a(b − c) = ab − ac .________________________________________________________
Let us examine the following part of the problem:
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→ " − 1(115 − 2x) " ;
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→ " − 1(115 − 2x) " = (-1 * 115) − (-1 * 2x) ;
= -115 − (-2x) ;
= -115 + 2x ;
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So we can bring down the: " {"B = 180 " ...}" portion ;
→and rewrite:
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→ B = 180 − 115 + 2x ;
→ B = 65 + 2x ;
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Now; given: "B = - 6x + 169 " ; ↔ B = 169 − 6x ;
→ " B = 169 − 6x = 65 + 2x " ;
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→ " 169 − 6x = 65 + 2x "
Subtract "65" from each side of the equation; & Subtract "2x" from each side of the equation:
→ 169 − 6x − 65 − 2x = 65 + 2x − 65 − 2x ;
to get:
→ " - 8x + 104 = 0 " ;
Subtract "104" from each side of the equation:
→ " - 8x + 104 − 104 = 0 − 104 " ;
to get:
→ " - 8x = - 104 ;
Divide each side of the equation by "-8" ;
to isolate "x" on one side of the equation; & to solve for "x" ;
→ -8x / -8 = -104 / -8 ;
to get:
→ x = 13 ;
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Now, let us solve for: " B " ; → {for which this very question/problem asks!} ;
→ B = 65 + 2x ;
Plug in our solved value, " 13 ", for "x" ;
→ B = 65 + 2(13) ;
= 65 + (26) ;
→ B = " 91 " .
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Also, check our answer:
_______________________________________________________
Given: "B = - 6x + 169 " ; ↔ B = 169 − 6x = 91 ;
When "x = 13 " ; does: " B = 91 " ?
→ Plug in our "solved value" of " 13 " for "x" ;
→ to see if: "B = 91" ; (when "x = 13") ;
→ B = 169 − 6x ;
= 169 − 6(13) ;
= 169 − (78)______________________________________________________
→ B = " 91 " .
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Answer:
0.5amps
Step-by-step explanation:
I is inversely proportional to resistance
I = k/R
k = I/ R
k = 20/5
k = 4
This means that, I = 4/R
So, when R = 8ohms,
I = 4/8
=0.5amps
The answer is false. They are the same