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givi [52]
3 years ago
12

What is the distance from X to Y?

Mathematics
1 answer:
densk [106]3 years ago
7 0

Answer:

15

Step-by-step explanation:

distance formula

\sqrt{((x2-x1)^2 + (y2-y1)^2)}

\sqrt{(9-0)^2 + (-6-6)^2}

\sqrt{81 + 144}

\sqrt{225}

15

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Please answer this question...
irina1246 [14]

Answer:  I believe the standard deviation will remain the same, 6, choice A.

Step-by-step explanation:  Standard deviation is calkculated based on the difference of each point from the mean.  If 10 were added to each score, the mean would also increase by 10.  That would mean that the difference between the mean and each score would be exactly the same, leading to an identical standard deviation.  Sounds crazy, but that's what numbers do for a living.

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3 years ago
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The angles of a quadrilateral are 140, 80, 60, and 80<br> What type of quadrilateral could it be?
My name is Ann [436]
Your answer is a trapezium 
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3 years ago
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an isosceles triangle has congruent sides of 20cm. the base is 10cm. find the height of the triangle.​
dolphi86 [110]

Answer:

\large\boxed{A_\triangle=25\sqrt{15}\ cm^2}

Step-by-step explanation:

Look at the picture.

The formula of an area of a triangle:

A_\triangle=\dfrac{bh}{2}

<em>b</em><em> - base</em>

<em>h</em><em> - height</em>

<em />

We need a length of a height.

Use the Pythagorean theorem:

leg^2+leg^2=hypotenuse^2

We have:

leg=5,\ leg=h,\ hypotenuse=20

Substitute:

5^2+h^2=20^2

25+h^2=400             <em>subtract 25 from both sides</em>

h^2=375\to h=\sqrt{375}\\\\h=\sqrt{(25)(15)}\\\\h=\sqrt{25}\cdot\sqrt{15}\\\\h=5\sqrt{15}\ cm

Calculate the area:

A_\triangle=\dfrac{(10)(5\sqrt{15})}{2}=\dfrac{50\sqrt{15}}{2}=25\sqrt{15}\ cm^2

8 0
3 years ago
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}\\\\&#10;c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;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}\\\\&#10;c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;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}\\\\&#10;c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;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
3 years ago
What is the cost of 20pens is rs 1200,what is the cost of one pen​
myrzilka [38]

Answer:

rs 60

1200/20=60......

3 0
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