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2 years ago

Use a graph to solve the equation 7x+3=9x-13 x=

1 answer:

3 0

Subtract 7x from both sides
3=9x-13-7x
simplify 9x-13-7x to 2x-13
3=2x-13
Add 13 to both sides
3+13=2x
simplify 3+13 to 16
16=2x
simplify both sides by 2
16/2=x
simplify16/2 to 8
x=8

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A farm packages eggs into cartons that hold two dozen eggs each. Last week, 5,398 eggs were collected. How many cartons will the

Sonbull [250]

Answer:

225

Step-by-step explanation:

We're told that each carton holds two dozen eggs, or 24 eggs. To find out how many cartons the farm will need, we divide the total amount of eggs by the amount of eggs that fit in each carton. So:

$\frac{5398}{24} = 224.916667$

Since the number is a decimal, you need to round up to make sure every egg is in a carton. So, your answer is 225.

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2 years ago

What is the quotient of 99 divided by 9081

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2 years ago

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yaroslaw [1]

Answer:

Step-by-step explanation:

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2 years ago

A customer calculated the cost of a new jacket, c, including a 7% sales

netineya [11]

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2 years ago

A metal beam was brought from the outside cold into a machine shop where the temperature was held at 65degreesF. After 5 min, t

ivolga24 [154]

Answer:

The beam initial temperature is 5 °F.

Step-by-step explanation:

If T(t) is the temperature of the beam after t minutes, then we know, by Newton’s Law of Cooling, that

$T(t)=T_a+(T_0-T_a)e^{-kt}$

where $T_a$ is the ambient temperature, $T_0$ is the initial temperature, $t$ is the time and $k$ is a constant yet to be determined.

The goal is to determine the initial temperature of the beam, which is to say $T_0$

We know that the ambient temperature is $T_a=65$ , so

$T(t)=65+(T_0-65)e^{-kt}$

We also know that when $t=5 \:min$ the temperature is $T(5)=35$ and when $t=10 \:min$ the temperature is $T(10)=50$ which gives:

$T(5)=65+(T_0-65)e^{k5}\\35=65+(T_0-65)e^{-k5}$

$T(10)=65+(T_0-65)e^{k10}\\50=65+(T_0-65)e^{-k10}$

Rearranging,

$35=65+(T_0-65)e^{-k5}\\35-65=(T_0-65)e^{-k5}\\-30=(T_0-65)e^{-k5}$

$50=65+(T_0-65)e^{-k10}\\50-65=(T_0-65)e^{-k10}\\-15=(T_0-65)e^{-k10}$

If we divide these two equations we get

$\frac{-30}{-15}=\frac{(T_0-65)e^{-k5}}{(T_0-65)e^{-k10}}$

$\frac{-30}{-15}=\frac{e^{-k5}}{e^{-k10}}\\2=e^{5k}\\\ln \left(2\right)=\ln \left(e^{5k}\right)\\\ln \left(2\right)=5k\ln \left(e\right)\\\ln \left(2\right)=5k\\k=\frac{\ln \left(2\right)}{5}$

Now, that we know the value of $k$ we can use it to find the initial temperature of the beam,

$35=65+(T_0-65)e^{-(\frac{\ln \left(2\right)}{5})5}\\\\65+\left(T_0-65\right)e^{-\left(\frac{\ln \left(2\right)}{5}\right)\cdot \:5}=35\\\\65+\frac{T_0-65}{e^{\ln \left(2\right)}}=35\\\\\frac{T_0-65}{e^{\ln \left(2\right)}}=-30\\\\\frac{\left(T_0-65\right)e^{\ln \left(2\right)}}{e^{\ln \left(2\right)}}=\left(-30\right)e^{\ln \left(2\right)}\\\\T_0=5$

so the beam started out at 5 °F.

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2 years ago