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2 years ago

A circle is the collection of points in a plane that are the same distance from the radius. True or false?

Mathematics

2 answers:

alexdok [17]2 years ago

8 0

My answer would be false. I hope this answers your question!

stira [4]2 years ago

7 0

<h2>Answer:</h2>

The given statement:

" A circle is the collection of points in a plane that are the same distance from the radius "

is a FALSE STATEMENT.

<h2>Step-by-step explanation:</h2>

Circle--

It is the collections of all the points in a plane which are at the same distance from a fixed point on a plane.

The fixed distance is called the radius of the circle.

and the fixed point on the plane is called a center of a circle.

The general equation of the circle is given by:

(x-h)²+(y-k)²=r²

where (h,k) is the center of the circle and r is the radius of the circle.

Hence, the statement is FALSE.

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I don't know if this is right... please someone help mee

worty [1.4K]

For the first circle, let's use the pythagorean theorem

$\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
c=\sqrt{8^2+15^2}\implies c=\sqrt{289}\implies c=17$

now, it just so happen that the hypotenuse on that triangle, is actually 17, but we used the pythagorean theorem to find it, and the pythagorean theorem only works for right-triangles.

so if the hypotenuse is actually 17, that means that triangle there is actually a right-triangle, meaning that the radius there, and the outside line there, are both meeting at a right-angle.

when an outside line touches the radius line, and they form a right-angle, the outside line is indeed a tangent line, since the point of tangency is always a right-angle with the radius.

now, let's check for second circle

$\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
c=\sqrt{11^2+14^2}\implies c=\sqrt{317}\implies c\approx 17.8044938$

well, low and behold, we didn't get our hypotenuse as 16 after all, meaning, that triangle is NOT a right-triangle, and that outside line is not touching the radius at a right-angle, therefore is NOT a tangent line.

let's check the third circle

$\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
c=\sqrt{33^2+56^2}\implies c=\sqrt{4225}\implies c=\stackrel{33+32}{65}$

this time, we did get our hypotenuse to 65, the triangle is a right-triangle, so the outside line is indeed a tangent line.

6 0

2 years ago

Noa hopes to pick the winning ticket during her math class’s raffle. the student who picks the ticket with a true statement will

Natali5045456 [20]

Answer: The winning ticket is $2(x-6)=-35$ is equivalent to $2x=-23$ .

Step-by-step explanation:

Let's check all the options

$4(x-5)=35$ is not equivalent to $4x=40$

$\text{since }4(x-5)=35\\\Rightarrow4x-20=35\\\Rightarrow4x=35+20\\\Rightarrow4x=55$

$8(x-7)=35$ is not equivalent to $8x=28$

$\text{since }8(x-7)=35\\\Rightarrow8x-56=35\\\Rightarrow8x=35+56\\\Rightarrow8x=91$

$2(x-6)=-35$ is equivalent to $2x=-23$

$\text{since }2(x-6)=-35\\\Rightarrow2x-12=-35\\\Rightarrow2x=-35+12\\\Rightarrow2x=-23$

$9(x-6)=-35$ is not equivalent to $9x=-29$

$\text{since }9(x-5)=-35\\\Rightarrow9x--40=-35\\\Rightarrow9x=-35+40\\\Rightarrow9x=5$

Thus, the winning ticket is $2(x-6)=-35$ is equivalent to $2x=-23$ .

4 0

2 years ago

Read 2 more answers

Find the solutions<br> -2x+8y=14 <br> -2x-2y=-26

emmasim [6.3K]

X=-33 y= 10

Leave first row the same multiply 2nd row by -1
-2x+8y=14
2x+2y=26
10y= 40
Y=4
-2x+8(10)=14
-2x+80=14

3 0

3 years ago

Select the equations that are equivalent to the equation y – 2 = 3(x + 6). Select all that apply.

amm1812

The correct answer for the exercise shown above is the first option and the las option, which are:

<span>y = 3x + 20
3x - y = -20
</span>
The explanation is shown below:

1. You have the folllowing expression given in the problem above:

y-2=3(x+6)

2. If you solve it for y, you have:

y-2=3x+18
y=3x+18+2
y=3x+20

3. And if you move the variables to the left member, you have:

3x-y=-20

Therefore, as you can see, the correct answer are the options mentioned above.

3 0

2 years ago

3. A model rocket is launched from the ground with an initial velocity of 352 ft/sec.

Hoochie [10]

Answer:

e. It will take 11 seconds to reach the maximum height of 1,936 feet.

f. It will take 22 seconds to return to the earth.

Step-by-step explanation:

Given:

Initial velocity $v_0$ = 352 ft/sec

Solving for question e.

To find the time required to reach the maximum height we will use the formula,

$h(t) = -16t^2+v_0t+h_0$ ,

where $v_0$ is the starting velocity

$h_0$ is the initial height.

Using the velocity and starting height from our problem we have,

$h(t) = -16t^2+352t+0$ ,

The path of this rocket will be a downward facing parabola, so there will be a maximum.

This maximum will be at the vertex of the graph.

To find the vertex we start out with $x= \frac{-b}{2a}$ which in our case is,

$x=\frac{-352}{2(-16)}=\frac{-352}{-32}= 11$

So, It will take 11 seconds for the rocket to reach its maximum height.

We will find maximum height using the formula by substituting value of t we get,

$h(11)=-16(11^2)+352(11)+0\\h(11) = -16 \times121+ 352 \times 11 = -1936+3872= 1936 \ ft$

Hence the maximum height will be $1936 \ ft$

Now Solving for question f.

To find the time required for rocket to reach earth.

We will set our formula to 0 to find the time.

$0= -16t^2+352t+0\\-16t(t-22)=0$

Using the zero product property, we know that either -16t = 0 or t - 22 = 0. When -16t = 0 is at t = 0, when the rocket is launched. t - 22 = 0 gives us an answer of t = 22.

So the rocket reaches the Earth again at 22 seconds.

8 0

2 years ago