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3 years ago

In order to inscribe a circle in a triangle, which line must you construct?

Mathematics

1 answer:

dalvyx [7]3 years ago

4 0

Answer:

You must construct an angle bisector. The point where two angle bisectors meet is the center of the circle.

The correct answer is D. angle bisector.

Hope this helps =)

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Show that f(2-3i) is a real number

densk [106]

Answer:

2 - 3i

is 2 - the square root of -9

2 -√-9

2 + √9

2 + 3 = 5

<em><u>Thank you (I dont know if these answers are 100% correct but I tried my best!)</u></em>

7 0

3 years ago

Factor a number, variable, or expression out of the trinomial 5x^2 - 25x + 10

Harman [31]

Hope you do well , i pray blessings and growth to you .

6 0

2 years ago

Find the volume of a prism. The area of the pentagonal base is (124/tan36°)m² . It’s height is 13 m. If necessary, round your an

lara [203]

Answer:

24.56

Step-by-step explanation:

5 0

3 years ago

OM , ON and OP are the angle bisectors of ∠AOB, ∠BOC, and ∠COD respectively. ∠AOD and ∠BOE are straight angles.

Orlov [11]

Find m∠BOC, if m∠MOP = 110°.

Answer:

m∠BOC= 40 degrees

Step-by-step explanation:

A diagram has been drawn and attached below.

OM bisects AOB into angles x and x respectively
ON bisects ∠BOC into angles y and y respectively
OP bisects ∠COD into angles z and z respectively.

Since ∠AOD is a straight line

x+x+y+y+z+z=180 degrees

$2x+2y+2z=180^\circ$

We are given that:

m∠MOP = 110°.

From the diagram

∠MOP=x+2y+z

Therefore:

x+2y+z=110°.

Solving simultaneously by subtraction

$2x+2y+2z=180^\circ$

x+2y+z=110°.

We obtain:

x+z=70°

Since we are required to find ∠BOC

∠BOC=2y

Therefore from x+2y+z=110° (since x+z=70°)

70+2y=110

2y=110-70

2y=40

Therefore:

m∠BOC= 40 degrees

3 0

3 years ago

Water is leaking out of a large barrel at a rate proportional to the square rooot of the depth of the water at that time. If the

sdas [7]

Answer:

It will take about 35.49 hours for the water to leak out of the barrel.

Step-by-step explanation:

Let $y(t)$ be the depth of water in the barrel at time $t$ , where $y$ is measured in inches and $t$ in hours.

We know that water is leaking out of a large barrel at a rate proportional to the square root of the depth of the water at that time. We then have that

$\frac{dy}{dt}=-k\sqrt{y}$

where $k$ is a constant of proportionality.

Separation of variables is a common method for solving differential equations. To solve the above differential equation you must:

Multiply by $\frac{1}{\sqrt{y}}$

$\frac{1}{\sqrt{y}}\frac{dy}{dt}=-k$

Multiply by $dt$

$\frac{1}{\sqrt{y}}\cdot dy=-k\cdot dt$

Take integral

$\int \frac{1}{\sqrt{y}}\cdot dy=\int-k\cdot dt$

Integrate

$2\sqrt{y}=-kt+C$

Isolate $y$

$y(t)=(\frac{C}{2} -\frac{k}{2}t)^2$

We know that the water level starts at 36, this means $y(0)=36$ . We use this information to find the value of $C$ .

$36=(\frac{C}{2} -\frac{k}{2}(0))^2\\C=12$

$y(t)=(\frac{12}{2} -\frac{k}{2}t)^2\\\\y(t)=(6 -\frac{k}{2}t)^2$

At t = 1, y = 34

$34=(6 -\frac{k}{2}(1))^2\\k=12-2\sqrt{34}$

So our formula for the depth of water in the barrel is

$y(t)=(6 -\frac{12-2\sqrt{34}}{2}t)^2\\\\y(t)=\left(6-\left(6-\sqrt{34}\right)t\right)^2\\$

To find the time, $t$ , at which all the water leaks out of the barrel, we solve the equation

$\left(6-\left(6-\sqrt{34}\right)t\right)^2=0\\\\t=3\left(6+\sqrt{34}\right)\approx 35.49$

Thus, it will take about 35.49 hours for the water to leak out of the barrel.

5 0

4 years ago