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3 years ago

Solve for x. Enter your answer below as an improper fraction in lowest terms, using the slash ( / ) as the fraction bar.

Mathematics

2 answers:

kykrilka [37]3 years ago

3 0

Divide 1/18 get the answer and do 3/4 and subtract by 3

wlad13 [49]3 years ago

3 0

I added 1/18 to 3 and 3/4 and got 137/36

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Does the relation y=x^2-1 defines y as a function of x

Reil [10]

Answer:

Yes.

From Definition - A special relationship where each input has a single output.

Check if there is going to be one y for one x.

Step-by-step explanation:

Re-write it in terms of y=…., meaning solve for y
If it looks like you have two equations then it is a relationship -case a
If you need to prove that one has a function for case b, you would point out that if you substitute any value you will get some value in Reals.

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3 years ago

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How many milliliters equal 1942 cm?

stiks02 [169]

1cm³=1mL
1942cm=x mL

1cm³ 1942cm³
________ = ________
1mL x mL

Cross multiply
1x=1942
divide both sides by 1
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Answer=1942mL

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3 years ago

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Jenna earns $7.75 per hour working at the mall. If Jenna worked 24 1/5 hours last week, how much did she earn before taxes were

Katyanochek1 [597]

24 1/5 is equal to 24.2

Multiply 24.2 by 7.75

24.2* 7.75 = 187.55

So she made $187.55 before taxes

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3 years ago

Dale made a scale drawing of a house and its lot. The scale he used was 1 inch: 7 feet. The actual length of the backyard is 35

Kipish [7]

Answer:

Step-by-step explanation:

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3 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.

6 0

3 years ago