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3 years ago

What is the solution to the inequality? −4(x−6)≤−2x+6

1 answer:

3 0

-4(x-6) ≤ - 2x+6
-4x-24≤ -2x+6
+2x +2x
-2x-24≤+6
+24 +24
-2x ≤ 30
Divide both sides by -2
and you get....
x ≥ -15
You switch the inequality sign because you're dividing by a negative.
I hope all is well, and you pass! Good luck, rockstar! (:

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An acute angle can have all of the following measures except 89 45 10 0

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2 years ago

How many zeros are in the standard form of 10 7 write this number in standard form

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Answer:

There are 7 zeroes in the standard form of 10⁷

The number is: 10,000,000

Explanation:

A power represents how many times this number is multiplied by itself

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x² means we will multiply x by itself 2 times (x * x)

x⁵ means we will multiply x by itself 5 times (x * x * x * x * x)

The standard form of a number means the number written with no powers

Now, the given number is 10⁷

This means that:

The number is 10 multiplied by itself 7 times

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Hope this helps :)

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3 years ago

The radius of a circle is 4 m. Find its area to the nearest whole number.

Slav-nsk [51]

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16π or around 50.26

Step-by-step explanation:

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2 years ago

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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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3 years ago