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KonstantinChe [14]
3 years ago
13

Which is the best description for the lines i and k?

Mathematics
1 answer:
Sonbull [250]3 years ago
6 0

Answer: intersecting but not perpendicular

Step-by-step explanation: Let's start by eliminating a few options.

First off, parallel lines are coplanar lines that do not intersect.

In the diagram shown, lines <em>i </em>and <em>k</em> do intersect

so option A doesn't make any sense.

We can also eliminate option C because by definition,

parallel lines <em><u>do not intersect </u></em> so we can cross this off.

We also can't conclude that lines <em>i</em> and <em>k </em>are perpendicular,

because we don't have a box at the vertex and from the

diagram shown, it doesn't look like they form a perfect right angle.

That leaves us with option D.

Notice that these two lines cross at a point.

When two lines cross at a point, we say that they intersect.

So they are intersecting lines but not perpendicular.

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Perfect Squaress less than 30​
levacccp [35]

The perfect whole squares are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.

Less than 30 are:

<h2>1, 4, 9, 16, 25</h2>
5 0
4 years ago
In New York, an Uber costs $15 plus $1.84 per mile. If you are traveling from the airport, there is an additional charge of $7.9
belka [17]

Answer: C; 12 miles

Solution:

1) Add the initial Uber cost ($15) and airport toll charge:

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Round down to 12 miles or option C

<h2 /><h2 />

3 0
3 years ago
Solve for 2.
Step2247 [10]

Answer:

(3x – 6)(-x + 3) = 0

=> 3x - 6 = 0 => x = 6/3 = 2

and

=> -x + 3 = 0 => x = 3

Hope this helps!

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5 0
3 years ago
What is 1 and 5/9 times 9
kolezko [41]
1 \frac{5}{9} * 9= 14

7 0
3 years ago
Read 2 more answers
An equilateral triangle is inscribed in a circle of radius 2r. Express the area A within the circle but outside the triangle as
defon

Answer:

A=8\pi r^2-\dfrac{49\sqrt{3}x^2}{4}\ un^2.

Step-by-step explanation:

An equilateral triangle with the side 7x units is inscribed in a circle of radius 2r units.

1. The area of the circle is

A_{circle}=2\pi (2r)^2=2\pi \cdot 4r^2=8\pi r^2\ un^2.

2. The area of the equilateral triangle is

A_{\triangle }=\dfrac{(7x)^2\sqrt{3}}{4}=\dfrac{49\sqrt{3}x^2}{4}\ un^2.

3. The area A within the circle but outside the triangle is

A=A_{circle}-A_{\triangle}=8\pi r^2-\dfrac{49\sqrt{3}x^2}{4}\ un^2.

7 0
3 years ago
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