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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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Maslowich

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no

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3 years ago
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Consider the parabola r​(t)equalsleft angle at squared plus 1 comma t right angle​, for minusinfinityless thantless thaninfinity
kodGreya [7K]

Given:-   r(t)=< at^2+1,t>  ; -\infty < t< \infty , where a is any positive real number.

Consider the helix parabolic equation :  

                                              r(t)=< at^2+1,t>

now, take the derivatives we get;

                                            r{}'(t)=

As, we know that two vectors are orthogonal if their dot product is zero.

Here,  r(t) and r{}'(t)  are orthogonal i.e,   r\cdot r{}'=0

Therefore, we have ,

                                  < at^2+1,t>\cdot < 2at,1>=0

< at^2+1,t>\cdot < 2at,1>=

                                              =2a^2t^3+2at+t

2a^2t^3+2at+t=0

take t common in above equation we get,

t\cdot \left (2a^2t^2+2a+1\right )=0

⇒t=0 or 2a^2t^2+2a+1=0

To find the solution for t;

take 2a^2t^2+2a+1=0

The numberD = b^2 -4ac determined from the coefficients of the equation ax^2 + bx + c = 0.

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Since, for any positive value of a determinant is negative.

Therefore, there is no solution.

The only solution, we have t=0.

Hence, we have only one points on the parabola  r(t)=< at^2+1,t> i.e <1,0>




                                               




6 0
3 years ago
Find the 8th term of the geometric sequence whose common ratio is 2/3 and whose first term is 7​
motikmotik

Answer:

The 8th term of the sequence is 896/2187.

Step-by-step explanation:

We want to find the 8th term of a geometric sequence whose common ratio is 2/3 and whose first term is 7.

We can write a direct formula. Recall that the direct formula of a geometric sequence is given by:

\displaystyle x_{n} = a\left(r\right)^{n-1}

Where <em>a</em> is the initial term and <em>r</em> is the common ratio.

Substitute:

\displaystyle x_{n} = 7\left(\frac{2}{3}\right)^{n-1}

To find the 8th term, let <em>n</em> = 8. Substitute and evaluate:

\displaystyle \begin{aligned} x_{8} &= 7\left(\frac{2}{3}\right)^{(8) - 1} \\ \\ &= 7\left(\frac{2}{3}\right)^{7} \\ \\ &= 7\left(\frac{128}{2187}\right) \\ \\ &= \frac{896}{2187} = 0.4096...\end{aligned}

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8 0
3 years ago
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x + 12 + 100 + x = 180

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Step-by-step explanation:

Given: Goal of running is at least 50 miles in a month.

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As given, Andrea have already run 7 miles.

∴ Andrea need to run this month= 50\ miles - 7\ miles= 43\ miles

Days remaining this month is 25.

Now, finding miles need to run each day to meet her goal.

Miles need to run each day= \frac{remaining\ miles}{remaining\ days}

⇒ Miles need to run each day= \frac{43}{25} = 1.72\ miles

∴ Andrea need to run at least 1.72 miles to meet her goal this month.

           

8 0
3 years ago
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