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Cerrena [4.2K]
3 years ago
7

PLEASEEEEE HELP ASAP WILL MARK BRAINIEST

Mathematics
1 answer:
Vinil7 [7]3 years ago
8 0

Answer:

Step-by-step explanation:

2 I think?

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Which statement about triangles is true
Salsk061 [2.6K]
C is the correct answer. An equilateral triangle is a type of triangle.
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I WILL GIVE YOU 40 POINTS AND THE HIGHEST ANSWER. YOU HAVE TO DO WHAT THE STEPS SAYS IN ORDER TO GET EVERYTHING
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This is a 30-60-90 triangle, in which the side opposite of the 90 degrees is the longest side (2), the side opposite of the 60 degrees is the √3 (2√3 in this case), and the last side (opposite of the 30 degrees) is the (1)

2√3/√3 = 2

So the side opposite  of the 30° = 2

a) the width is 2 miles

The length of the side walk is 2(a)
a = 2

2(2) = 4

b) The length of the sidewalk is 4 miles

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5 0
3 years ago
PLEASE HELP. <br> if v1….
Rainbow [258]
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6 0
2 years ago
find the angle between the vectors. (first find the exact expression and then approximate to the nearest degree. ) a=[1,2,-2]. B
SashulF [63]

Answer:

\theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190

Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel

And the anfle is approximately \theta \approx 48

Step-by-step explanation:

For this case first we need to calculate the dot product of the vectors, and after this if the dot product is not equal to 0 we can calculate the angle between the two vectors in order to see if there are parallel or not.

a=[1,2,-2], b=[4,0,-3,]

The dot product on this case is:

a b= (1)*(4) + (2)*(0)+ (-2)*(-3)=10

Since the dot product is not equal to zero then the two vectors are not orthogonal.

Now we can calculate the magnitude of each vector like this:

|a|= \sqrt{(1)^2 +(2)^2 +(-2)^2}=\sqrt{9} =3

|b| =\sqrt{(4)^2 +(0)^2 +(-3)^2}=\sqrt{25}= 5

And finally we can calculate the angle between the vectors like this:

cos \theta = \frac{ab}{|a| |b|}

And the angle is given by:

\theta = cos^{-1} (\frac{ab}{|a| |b|})

If we replace we got:

\theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190

Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel

And the anfle is approximately \theta \approx 48

3 0
3 years ago
Write 9 41/80 as a decimal
jolli1 [7]
Leave the whole number alone for now. Divide 41 by 80 and you get 0.5125

So your answer is 9.5125
6 0
3 years ago
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