Which of the following statements are true about line segments ??

What is the following equation formed when completing the square on Y^2-12y=-27

Salsk061 [2.6K]

Answer:y^1=3 ,y^2=9 if im wrong sorry im wrong but hope you get an a

7 0

3 years ago

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If r = 4 units and x = 13 units, then what is the volume of the cylinder

irina [24]

Im going to go with the volume being 653.45 because v = pie r squared h = pie times 4 squared times 13 which equals 653.45... hope its correct !

6 0

3 years ago

Help me with this question please, explain why u choose that answer and show ur steps

Setler [38]

Let’s condense the original expression and see if that’s one of the answers

-1/2(-3/2 x + 6x + 1) - 3x

First we distribute the -1/2

((-1/2)(-3/2 x) + (-1/2)(6x) + (-1/2)(1)) -3x

3/4x - 3x - 1/2 -3x

Combine like terms

(3/4 x - 3x - 3x) -1/2

-21/4 x - 1/2

We could make this a mixed number

-5 1/4 x - 1/2

(Mixed number: 5 times 4 plus one gets the numerator)

So the answer is D

7 0

3 years ago

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the perpendicular distance of a point from the x-axis is 4 units and the perpendicular distance from the y axis is 5 units write

quester [9]

Step-by-step explanation:

let p be the point which is at a distance of 4 units from the x-axis and at a distance of 5 units from the y axis

(i) When P lies in the first quadrant then the coordinates of P are (5,4).

(ii) When of P lies in the second quadrant then the coordinates of P are (-5,4).

(iii) When P lies in the third quadrant then the coordinates of P are (-5,-4).

(iv) When P lies in the fourth quadrant then the coordinates of P are (5,-4).

<h3>Hope it helps you!!</h3>

7 0

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.

6 0

3 years ago