-4 > -9
-4 is closer to 0 than -9
it is further right on the number line
Answer:
See below ~
Step-by-step explanation:
<u>Circle 1</u>
- Circumference = 2πr = 2 x 3.14 x 2.5 = 5 x 3.14 = 15.7 m
- Area = πr² = 3.14 x (2.5)² = 3.14 x 6.25 = 19.625 m²
<u>Circle 2</u>
- Area = πr² = 3.14 x 1² = 3.14 mm²
- Circumference = 2πr = 2 x 3.14 x 1 = 6.28 mm
<h2>
Introduction</h2>
Hello! I hope you are having a nice day. My name is Galaxy and I will be helping you with this problem. We can solve this problem in 2 steps, respectively Theory and Solving.
I'll go ahead and start with the Theory.
<h2>
Theory</h2>
Before we attempt to solve the problem mathematically, we must first figure out how we're going to solve this problem.
We know that we have a line and a point, we can start by graphing the equation and point that we've received.
<h2 /><h2>
Solving</h2>
Now that we have our points plotted and our equations graphed, we can start to see that something odd is happening, the given point is on the line itself.
We can check this by inputting the points into our equation:

This proves that our point is on the line that we were given.
According to the basic rules of Euclidean Geometry, we know that parallel lines can never intersect each other and due to this there cannot be any solution to this problem.
Any other line using this point would intersect the given line, and due to this, this problem has no real solutions.
* The only line that can use these points and graph is the line provided, and that cannot work due to the lines intersecting at an infinite number of points.
Cheers,
Galaxy.
Answer:
x = 7.5
Step-by-step explanation:
We first labelled the diagram we get
DC = 8
AD = 10
CE = 6
BE = x
∴ AC =AD + DC = 18
∴ BC = BE + ED = x + 6
To Find :
BE = x =?
Solution:
Let DE || AC
In Δ ABC and Δ DEC
∠A ≅ ∠D …………..{corresponding angles ∵ DE || AB }
∠B ≅ ∠E ..............{corresponding angles ∵ DE || AB }
∠C ≅ ∠C ……….....{Reflexive Property}
Δ ABC ~ Δ DEC ….{Angle-Angle-Angle Similarity test}
If two triangles are similar then their sites are in proportion.
On substituting the given values we get
∴ 