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ki77a [65]
2 years ago
11

candy bars have increased in price by 8% in recent years. if today’s price is approximately $1.60, how much was the original pri

ce of candy bars before the price increase?
Mathematics
2 answers:
igor_vitrenko [27]2 years ago
5 0

Answer:

$1.50

Step-by-step explanation:

frez [133]2 years ago
5 0

Answer:

1.48

Step-by-step explanation:

no step by step explanation

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Please answer quick!
inessss [21]
The answer should be the 3rd one
5 0
3 years ago
Visitors to a carnival can buy an unlimited-ride pass for 50$ or an entrance-only pass for 20$. In one day, 282 passss were sold
olchik [2.2K]
20*282= 5640
10680-5640=5040
50-20=30
5040 dividend by 30= 168
Answer: 168
6 0
3 years ago
The number N of cars produced at a certain factory in 1 day after t hours of operation is given by Upper N (t )equals 800 t minu
Elden [556K]

Answer:

The cost C as a function of t is C(t) = 30,000 + 6,400,000 t - 40,000 t²

Step-by-step explanation:

The function N(t) = 800 t -  5t², represents the number of cars produced at a time t hours in a day, where 0 ≤ t ≤ 10

The function C(N) = 30,000 + 8,000 N, represents the cost C​ (in dollars) of producing N cars

We need to find The cost C as a function of the time t

That means Substitute N in C by its function by other word find the composite function (C о N)(t)

∵ C(N) = 30,000 + 8,000 N

∵ N(t) = 800 t - 5 t²

- Substitute N in C by 800 t - 5 t²

∴ C(N(t)) = 30,000 + 8000(800 t - 5 t²)

- Multiply the bracket by 8000

∴ C(N(t)) = 30,000 + 8000(800 t) - 8000(5 t²)

∴ C(N(t)) = 30,000 + 6,400,000 t - 40,000 t²

- C(N(t) = C(t)

∴ C(t) = 30,000 + 6,400,000 t - 40,000 t²

The cost C as a function of t is C(t) = 30,000 + 6,400,000 t - 40,000 t²

5 0
3 years ago
Solve the proportion
mario62 [17]

Answer:

\boxed{\sf x = 5}

Step-by-step explanation:

\sf Solve  \: for  \: x  \: over  \: the  \: real \:  numbers:  \\ \sf \implies  \frac{2}{x - 3}   =  \frac{5}{x}  \\  \\  \sf Take  \: the \:  reciprocal  \: of  \: both \:  sides:  \\ \sf \implies  \frac{x - 3}{2}  =  \frac{x}{5}  \\  \\  \sf Expand  \: out \:  terms \:  of \:  the \:  left  \: hand \:  side:  \\  \\ \sf \implies \frac{x}{2}  -  \frac{3}{2}  =  \frac{x}{5}  \\  \\  \sf Subtract \:  \frac{x}{5}   -  \frac{3}{2}  \: from  \: both  \: sides: \\  \sf \implies \frac{x}{2}  -  \frac{3}{2} - ( \frac{x}{5}   -  \frac{3}{2} ) =  \frac{x}{5} - ( \frac{x}{5}  -  \frac{3}{2} ) \\  \\  \sf \implies \frac{x}{2}  -  \frac{3}{2} -  \frac{x}{5}    +   \frac{3}{2} =  \frac{x}{5} -  \frac{x}{5}  +  \frac{3}{2}  \\  \\  \sf \frac{x}{5}  -  \frac{x}{5}  = 0 :  \\  \sf \implies \frac{x}{2}  -  \frac{x}{5}  -  \frac{3}{2}  +  \frac{3}{2}  =  \frac{3}{2}  \\  \\  \sf  \frac{3}{2}   -   \frac{3}{2}   = 0:  \\  \sf \implies \frac{x}{2}  -  \frac{x}{5}  =  \frac{3}{2}   \\  \\ \sf \frac{x}{2}  -  \frac{x}{5} =  \frac{5x - 2x}{10}  =  \frac{3x}{10} :  \\   \sf \implies \frac{3x}{10}  =  \frac{3}{2}   \\  \\ \sf Multiply \:  both  \: sides \:  by \:  \frac{10}{3}  : \\   \sf \implies \frac{3x}{10}  \times  \frac{10}{3}  =  \frac{3}{2 }  \times  \frac{10}{3}   \\  \\ \sf \frac{3x}{10}  \times  \frac{10}{3}  =   \cancel{\frac{3}{10} } \times( x) \times  \cancel{ \frac{10}{3} } = x :  \\  \sf \implies x =  \frac{3}{2}  \times  \frac{10}{3} \\  \\   \sf  \frac{3}{2}  \times  \frac{10}{3}  = \cancel{ \frac{3}{2} }  \times \cancel{ \frac{3}{2} }  \times 5 :   \\ \sf \implies x = 5

8 0
3 years ago
The burges are moving across the country .Mr burges leaves 3 hours before mrs.burges. If he averages 50 mph and she averages 70
kirill [66]

10.5 hours

let t be time then distance travelled by both is

Mr Burges ⇔ d = 50t ( distance = speed × time )

Mrs Burges ⇒ d = 70(t - 3)

At the point of overtaking both will have travelled the same distance

equating their distance and solving for t

70(t - 3) = 50t

70t - 210 = 50t ( subtract 50t from both sides )

20t - 210 = 0 ( add 210 to both sides )

20t = 210 ( divide both sides by 20 )

t = 10.5 hours


4 0
3 years ago
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