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Vilka [71]
3 years ago
5

|2x+1|>5 what is the answer​solve for real numbers x.

Mathematics
1 answer:
umka2103 [35]3 years ago
4 0

Answer:

see explanation

Step-by-step explanation:

Inequalities of the type | x | > a always have solutions of the form

x < - a or x > a, thus

2x + 1 < - 5 OR 2x + 1 > 5 ( subtract 1 from both sides of both inequalities

2x < - 6 OR 2x > 4 ( divide both sides of both inequalities by 2 )

x < - 3 OR x > 2

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N + 7 &lt; -3<br><br> solve and show work please
Hitman42 [59]

Answer:

n < -10

Step-by-step explanation:

In order to get the variable n by itself subtract 7 from both sides across the inequality

n + 7 < -3

  -7     -7

n   <  -10

4 0
3 years ago
Fresh Fruits sells 10 lemons for $2.50. What is the price for 1 lemon
Mnenie [13.5K]

Answer:

25 cents per lemon

Step-by-step explanation:

Simply divide

7 0
3 years ago
Read 2 more answers
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
Tina was given the function y = x 2+ x – 5. For what x value(s) would y = 15?
ziro4ka [17]

Answer:

the correct answer is 4

Step-by-step explanation:

y = x^2 + x - 5

y = 4^2 + 4 - 5

y= 16 + 4 -5

y = 15

6 0
3 years ago
Multiply the terms 3xyz and -5x²yz​
Evgesh-ka [11]

Answer:

-15X^3y^2Z^2 is the answer

3 0
3 years ago
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