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tester [92]
3 years ago
6

Kimberly sells handmade jewelry online, charging one price for bracelets and another price for necklaces. This table shows how m

any bracelets and necklaces she sold in November and December.
Bracelets Necklaces Total sales

November 15 8 $500

December 19 24 $980

What was the price of each item?

Bracelets cost $____

each and necklaces cost $____
Mathematics
1 answer:
kenny6666 [7]3 years ago
5 0
Bracelet cost = b
Necklace cost = n

15b + 8n = 500
19b + 24n = 980

Elimination method x -3
-45b - 24n = -1500
19b + 24n = 980

-26b = -520
b = $20 bracelets

Plug back in
19(20) + 24n = 980
380 + 24n = 980
24n = 600
n = $25 necklaces


Check:
15(20)+ 8(25)= 500
300 + 200 = 500
500 = 500 :)
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A sofa is 900 more than the price of c which system of equation can be used to find price of each piece
Doss [256]

Answer:

$1200

Step-by-step explanation:

 

S + C = 1200

 

S = C + 900

 

C + 900 + C = 1200

 

2C + 900 = 1200

 

2C = 300

 

C = 150

 

The chair costs $150. The sofa is $1050.

Together they cost $1200.

5 0
3 years ago
Please help me with this thanks
Ronch [10]

If x = 1, then 3*1 = 3 which when modded with 5, we get 3 as a remainder. In other words, 3/5 = 0 remainder 3. We don't use the quotient at all when it comes to modular arithmetic. All we care about is the remainder.

If x = 2, then 3*2 = 6 which leads to remainder 1 when we divide by 5. Therefore, 3x = 1 (mod 5) when x = 2.

If x = 3, then 3*3 = 9 = 4 (mod 5) because 9/5 = 1 remainder 4.

So 3x = 4 (mod 5) when x = 3.

<h3>The final answer is C) 3</h3>

We don't need to check D since x = 3 is a solution and it's smaller than x = 4.

If you wanted to check x = 4, then 3*4 = 12 = 2 (mod 5) because 12/5 yields a remainder of 2.

7 0
3 years ago
G=x-c/x. for x<br><br>how do yoy make this an equation for x? help
zzz [600]
1.
\displaystyle \text{if }\quad G=x-  \frac{C}{x} \\\\

x \neq 0

Multiply both sides by x:

\displaystyle Gx=x\bigg(x- \frac{C}{x}\bigg )\\\\Gx=x^2-C

Subtract from both sides Gx:
Gx-Gx=x^2-C-Gx\\x^2-Gx-C=0

Then solve this equation:

\displaystyle x^2-Gx-C=0\\\\x_{1,2}= \frac{G\pm  \sqrt{G^2+4C} }{2}

2.
\displaystyle\text{if}\quad G= \frac{x-C}{x}

x \neq 0

Multiply both sides by x:

Gx=x-C

Subtract from both sides Gx:

x-C-Gx=0

Then solve this equation:

\displaystyle x-Gx-C=0\\x(1-G)=C\\x= \frac{C}{1-G}




7 0
3 years ago
Two bikers are riding a circular path.
kolbaska11 [484]

<h2>\huge \mathfrak \orange{Question}</h2>

Two bikers are riding a circular path.

The first rider completes a round in 12

minutes. The second rider completes

a round in 18 minutes. If they both

started at the same place and time

and go in the same direction, after

how many minutes will they meet

again at the starting point?

<h2>\huge \mathfrak \orange{Solution}</h2>

\huge \mathsf \blue{Given}

  • First rider takes 12 minutes to complete a round.
  • Second rider takes 18 minutes to complete a round.

\huge \mathsf \blue{To\: Find}

After how many minutes will they meet

again at the starting point?

Take the LCM of 12 and 18

12 = 2 × 2 × 3

18 = 2 × 3 × 3

Thus, the LCM of 12 and 18 is 36.

<h3>So they will meet after 36 minutes again at the starting point.</h3>
8 0
3 years ago
Read 2 more answers
Can someone plss help me with this question!!
Ganezh [65]

Answer: B

1 Newton with a force going right will remain

In D the object wont move

In A the object wont move

In C the object will move left

7 0
3 years ago
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