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2 years ago

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6,000 people voted in a town election. Of the two candidates, the winning candidate received 4,000 more votes than the losing ca

ndidate. How many votes did the winner get?

1 answer:

AVprozaik [17]2 years ago

8 0

Answer:

Step-by-step explanation:

let x = the number of votes of the first person

winner = x + 4000

x+ x+4000 = 6000

2x=2000

x=1000

first person = x = 1000

winner = x+4000 = 5000

The winner got 5000 votes

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Find the value of x (4x-9) (3x+29)

polet [3.4K]

Answer:

12x³+89x²-261x (?)

Step-by-step explanation:

For this equation, I am going to assume you want to find the value of the expression and you're multiplying all the values (since you can't find the value of x with this unless you had an equation).

(x)(4x-9)(3x+29)

(4x²-9x)(3x+29)

12x³+89x²-261x

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3 years ago

Solve the system of equations: 10x + 7y = 99 and y – x = 2.

svetlana [45]

7y + 10x = 99
y - x = 2
y = 7 , x = 5

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3 years ago

7.Suppose that water is entering a cylindrical water tank so that the initial height of the water is 3 cm and the

lukranit [14]

Answer:

$H(t) = 3*2^{t/30}$

Step-by-step explanation:

Height doubles at every 30 seconds, so let's call n how many times it happens, and the t the total time:

n = t/30

If the height doubles one time, it is multiplied by 2, if it doubles 2 times, it's multiplied by 2², for n times, it is multiplied by 2ⁿ, so for an initial height equal to 3:

$H(t) = 3*2^{t/30}$

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2 years ago

An ac unit uses 3500 watts or power. If it runs for 10 hours each day and the utility company charges $0.13 per kWh, how much do

Helga [31]

Cost = (3500 W)·(1 kw/1000 W)·(10 hrs/day)·(30 day/month)·($0.13/kw-hr)

Cost = (3500·1·10·30·0.13 / 1000)·(W·kW·hr·da·$ / W·da·month·kw·hr)

Cost = <em> 136.50 $/month</em>

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3 years ago

Situation:A researcher in North America discoversa fossile that contains 65% of its originalamount of C-14..-ktN=NoeNo inital am

zheka24 [161]

SOLUTION

We have been given the equation of the decay as

$\begin{gathered} N=N_0e^{-kt} \\ where\text{ } \\ N_0=initial\text{ amount of C-14 at time t} \\ N=amount\text{ of C-14 at time t = 65\% of N}_0=0.65N_0 \\ k=0.0001 \\ t=time\text{ in years = ?} \end{gathered}$

So we are looking for the time

Plugging the values into the equation, we have

$\begin{gathered} N=N_0e^{-kt} \\ 0.65N_0=N_0e^{-0.0001t} \\ e^{-0.0001t}=\frac{0.65N_0}{N_0} \\ e^{-0.0001t}=0.65 \end{gathered}$

Taking Ln of both sides, we have

$\begin{gathered} ln(e^{-0.000t})=ln(0.65) \\ -0.0001t=ln(0.65) \\ t=\frac{ln(0.65)}{-0.0001} \\ t=4307.82916 \end{gathered}$

Hence the answer is 4308 to the nearest year

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8 months ago