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Margarita [4]
3 years ago
13

last week a candle store recieved 335.60 for selling 20 candles. small candles sell for 10.98 and largr candles sell fer 27.98.

how many large canfles did the store sell
Mathematics
1 answer:
mina [271]3 years ago
5 0
Hmmmm closest I can get to this answer is 6 or 7. The numbers don't add up and it's probably why you haven't received an answer. If you do this math:
13small candles multiplied by 10.98 = 142.74
7large candles multiplied by 27.98 = 195.86
142.74+195.86 = 338.60 which is OVER the total received.
7+13=20 candles
14small candles multiplied by 10.98 = 153.72
6 large candles multiplied by 27.98 = 167.88
153.72+167.88 = 321.60 which is UNDER the total received
Can't go any lower or higher because it's even lower or higher. 
6+14=20 candles
So I wouldn't really know what to say. This isn't adding up.

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AD is the perpendicular bisector of BC.
masha68 [24]

5x-10= 3x+10

5x-3x= 10+10

2x÷2= 20÷2

x= 10

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3 years ago
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What is the value of a in the equation 3a+b=54, when b=9?
insens350 [35]

3a+9=54

3a+9-9=54-9

3a= 45

a=15

Check answer:

3(15)+9=54

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Answer is a=15

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3 years ago
5u^2 + 6 = -13u please help
Komok [63]

Answer:

u = -2, -3/5

General Formulas and Concepts:

<u>Algebra I</u>

  • Standard Form: ax² + bx + c = 0
  • Factoring
  • Finding roots

Step-by-step explanation:

<u>Step 1: Define equation</u>

5u² + 6 = -13u

<u>Step 2: Solve for </u><em><u>u</u></em>

  1. Rewrite:                    5u² + 13u + 6 = 0
  2. Factor:                      (u + 2)(5u + 3) = 0
  3. Find roots:                u = -2, -3/5
5 0
2 years ago
The number line shows the graph of inequality.​
Sophie [7]

Answer:

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Step-by-step explanation:

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I hope this helps.

4 0
2 years ago
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the half life of c14 is 5730 years. Suppose that wood found at an archeological excavation site contains about 35% as much C14 a
Furkat [3]

Answer:

The wood was cut approximately 8679 years ago.

Step-by-step explanation:

At first we assume that examination occured in 2020. The decay of radioactive isotopes are represented by the following ordinary differential equation:

\frac{dm}{dt} = -\frac{m}{\tau} (Eq. 1)

Where:

\frac{dm}{dt} - First derivative of mass in time, measured in miligrams per year.

\tau - Time constant, measured in years.

m - Mass of the radioactive isotope, measured in miligrams.

Now we obtain the solution of this differential equation:

\int {\frac{dm}{m} } = -\frac{1}{\tau}\int dt

\ln m = -\frac{1}{\tau} + C

m(t) = m_{o}\cdot e^{-\frac{t}{\tau} } (Eq. 2)

Where:

m_{o} - Initial mass of isotope, measured in miligrams.

t - Time, measured in years.

And time is cleared within the equation:

t = -\tau \cdot \ln \left[\frac{m(t)}{m_{o}} \right]

Then, time constant can be found as a function of half-life:

\tau = \frac{t_{1/2}}{\ln 2} (Eq. 3)

If we know that t_{1/2} = 5730\,yr and \frac{m(t)}{m_{o}} = 0.35, then:

\tau = \frac{5730\,yr}{\ln 2}

\tau \approx 8266.643\,yr

t = -(8266.643\,yr)\cdot \ln 0.35

t \approx 8678.505\,yr

The wood was cut approximately 8679 years ago.

5 0
3 years ago
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