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Elenna [48]
4 years ago
5

HELP ASAP WITH THIS QUESTION!!!

Mathematics
2 answers:
amid [387]4 years ago
8 0
That would be the bottom left one....because all of the others are spaced evenly
Jet001 [13]4 years ago
6 0
The answer is the bottom left
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How many unique triangles can be drawn with side lengths 8 in., 12 in., and 24 in.? Explain.
Elanso [62]

Answer:

2

Step-by-step explanation:

To figure this out, you must use the triangle inequality theorem. It says that the sum of 2 sides of the triangle should be more than the third side. If you add 8 and 12, you get less than 24 but if you add 12 and 24 or 8 and 24, you will get more. So, there are only 2 triangles! I hope this helps!

8 0
3 years ago
Find the direction cosines and direction angles of the vector. (Give the direction angles correct to the nearest degree.) 5, 1,
Dahasolnce [82]

Answer:

The direction cosines are:

\frac{5}{\sqrt{42} }, \frac{1}{\sqrt{42} }  and  \frac{4}{\sqrt{42} }  with respect to the x, y and z axes respectively.

The direction angles are:

40°,  81° and  52° with respect to the x, y and z axes respectively.

Step-by-step explanation:

For a given vector a = ai + aj + ak, its direction cosines are the cosines of the angles which it makes with the x, y and z axes.

If a makes angles α, β, and γ (which are the direction angles) with the x, y and z axes respectively, then its direction cosines are: cos α, cos β and cos γ in the x, y and z axes respectively.

Where;

cos α = \frac{a . i}{|a| . |i|}               ---------------------(i)

cos β = \frac{a.j}{|a||j|}               ---------------------(ii)

cos γ = \frac{a.k}{|a|.|k|}             ----------------------(iii)

<em>And from these we can get the direction angles as follows;</em>

α =  cos⁻¹ ( \frac{a . i}{|a| . |i|} )

β = cos⁻¹ ( \frac{a.j}{|a||j|} )

γ = cos⁻¹ ( \frac{a.k}{|a|.|k|} )

Now to the question:

Let the given vector be

a = 5i + j + 4k

a . i =  (5i + j + 4k) . (i)

a . i = 5         [a.i <em>is just the x component of the vector</em>]

a . j = 1            [<em>the y component of the vector</em>]

a . k = 4          [<em>the z component of the vector</em>]

<em>Also</em>

|a|. |i| = |a|. |j| = |a|. |k| = |a|           [since |i| = |j| = |k| = 1]

|a| = \sqrt{5^2 + 1^2 + 4^2}

|a| = \sqrt{25 + 1 + 16}

|a| = \sqrt{42}

Now substitute these values into equations (i) - (iii) to get the direction cosines. i.e

cos α = \frac{5}{\sqrt{42} }

cos β =  \frac{1}{\sqrt{42} }              

cos γ =  \frac{4}{\sqrt{42} }

From the value, now find the direction angles as follows;

α =  cos⁻¹ ( \frac{a . i}{|a| . |i|} )

α =  cos⁻¹ ( \frac{5}{\sqrt{42} } )

α =  cos⁻¹ (\frac{5}{6.481} )

α =  cos⁻¹ (0.7715)

α = 39.51

α = 40°

β = cos⁻¹ ( \frac{a.j}{|a||j|} )

β = cos⁻¹ ( \frac{1}{\sqrt{42} } )

β = cos⁻¹ ( \frac{1}{6.481 } )

β = cos⁻¹ ( 0.1543 )

β = 81.12

β = 81°

γ = cos⁻¹ ( \frac{a.k}{|a|.|k|} )

γ = cos⁻¹ (\frac{4}{\sqrt{42} })

γ = cos⁻¹ (\frac{4}{6.481})

γ = cos⁻¹ (0.6172)

γ = 51.89

γ = 52°

<u>Conclusion:</u>

The direction cosines are:

\frac{5}{\sqrt{42} }, \frac{1}{\sqrt{42} }  and  \frac{4}{\sqrt{42} }  with respect to the x, y and z axes respectively.

The direction angles are:

40°,  81° and  52° with respect to the x, y and z axes respectively.

3 0
3 years ago
Which expression means “ multiply 3 by the sum of m and p “
charle [14.2K]

Answer:

3m+3p

Step-by-step explanation:

Sum of m and p is expressed as;

m + p (sum means addition)

If 3 is multiplied  by the sum, this is expressed as;

3(m+p)

Expand

3m + 3p

Hence the required expression is 3m+3p

8 0
3 years ago
Given f(x) and g(x) = f(x) + k, look at the graph below and determine the value of k.
Anton [14]

Answer:

Given the graph f(x) = \frac{1}{3}x -2 and g(x) = \frac{1}{3}x + 3

We have to find the value of k;

Since, g(x) = f(x) +k

\frac{1}{3}x+3 = \frac{1}{3}x-2 +k

Subtract \frac{1}{3}x from both sides we get;

3 = -2 +k

Add 2 to both sides we get;

3+2 = -2 +k +2

Simplify:

5 = k

or

k = 5

Therefore, the value of k = 5


3 0
3 years ago
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monitta
You have to pay $250.56 in taxes 
4 0
3 years ago
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