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AleksAgata [21]
2 years ago
15

What is the distance between the points (4, 7) and (4, −5)?

Mathematics
1 answer:
PIT_PIT [208]2 years ago
3 0

Answer:

12 units

Step-by-step explanation:

hope this helped luv!

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PLease Help! I will give you the brainiest and a lot of points!
gtnhenbr [62]

Answer:

this is the venn diagram I guess..

Step-by-step explanation:

find answer from this.. and check if it is correct

6 0
2 years ago
PLZ HELP AND SHOW YOUR WORK &lt;3333<br><br><br> Substitution.<br><br> 3x+2y=11<br><br> y=5x-1
liraira [26]
Hey there! :D

Substitute y for 5x-1. 

3x+2y=11

y=5x-1

3x+2(5x-1)=11

3x+10x-2=11

13x-2=11

13x=13

x=1

Now, plug that in to find y.

y=5x-1

y=5(1)-1

y=5-1

y=4

(1,4) <== the solution

I hope this helps!
~kaikers
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
Will give brainiest for this
cluponka [151]

Answer:

lol, u can still ask the question

Step-by-step explanation:

3 0
2 years ago
What is the length of line segment KJ?<br><br> 2√3 units<br> 3√2units<br> 3√3units<br> 3√5units
Pavlova-9 [17]
KJ^2=6^2+3^2
KJ=the square root of 36+9
KJ=square root of 45
kJ= 3 times the square root of 5
The answer is D
3 0
3 years ago
Read 2 more answers
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