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V125BC [204]
2 years ago
14

While visiting the 200, Elijah, Alda, and Chris create a triangle by standing at its vertices. They are the same distance away f

rom the snake exhibit. Which point of concurrency describes the location of the snake exhibit in relationship to
Elijah Aida, and Chris?
HELP PLEASE!
Mathematics
1 answer:
trapecia [35]2 years ago
5 0

Answer:

  • Circumcenter

Step-by-step explanation:

The circumcenter of a triangle is the point that has same distance from the three vertices of the triangle.

The snake exhibit is in the circumcenter of the triangle.

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dexar [7]

Answer:

1/5 . 950

This is cause 20% =20/100 and when reduced it is 1/5

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3 years ago
Given 2x - 4y = 24, the x-intercept is 12 and the y-intercept is -6.
Masteriza [31]

Answer:

False

Step-by-step explanation:

1. Plug in the numbers

2(x)-4(y)=24 -> 2(12)-4(-6)=24

2. Solve

2x12=24 & -4x(-6)=24

24-24=0

0≠24

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The volume of the cylinder
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15550.88

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Evaluate the cosine angle X please !
sveta [45]

Answer:

in right angled triangle

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5 0
3 years ago
Find the measure of the angle between the vectors 〈6, –3, 1〉 and 〈8, 9, –11〉
Pepsi [2]

Answer:

Option C.

Step-by-step explanation:

We know that for two vectors:

A = 〈a₁, a₂, a₃ 〉and  B =〈b₁, b₂, b₃〉.

The dot product between A and B is:

A.B = IAI*IBI*cos(θ)

Where θ is the angle between the vectors.

And the dot product can be also written as:

A.B = a₁*b₁ + a₂*b₂ + a₃*b₃

With that, we can find the angle between our vectors.

In this case, the vectors are:

A = 〈6, –3, 1〉 and B = 〈8, 9, –11〉

The modules are:

IAI = √( 6^2 + (-3)^2 + 1^2) = 6.78

IBI = √( 8^2 + 9^2 + (-11)^2) = 16.31

And the dot product between A and B is:

A.B = 6*8 + (-3)*9 + 1*(-11) = 10

Then we have that:

10 = IAI*IBI*cos(θ) = ( 6.78)*(16.31)*cos(θ)

10/(6.78*16.31) = cos(θ)

Now remember the inverse cosine function, Acos(x), this function has the property:

Acos(cos(x)) = x

If we apply this to both sides, we get:

Acos(10/(6.78*16.31)) = Acos(cos(θ)) = θ = 84.8°

The correct option is C.

7 0
3 years ago
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