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Bess [88]
2 years ago
9

Which values are in the solution set of the compound inequality? Select two options

Mathematics
1 answer:
zhannawk [14.2K]2 years ago
7 0

The values are in the solution set of the compound inequality are (-6, -3), (3, 8).

<h3>What is inequality?</h3>

Inequality is the phenomenon of unequal distribution of resources and opportunities among the given function.

To find the number that are solutions to the compound inequalities, the value of x is

First inequality:

4(x +3) ≤0

4x +12 ≤0

4x ≤-12

x ≤-3

The value of x lies in the interval = (-∞ , -3]

Second inequality:

x +1 >3

x>2

The value of x lies in the interval = (2 , ∞)

So, the interval solution is (-∞ , -3] U (2 , ∞)

Thus, the values are in the solution set of the compound inequality are (-6, -3), (3, 8).

Learn more about inequality.

brainly.com/question/20383699

#SPJ1

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Fine the exact value. <br> tan(60°)+tan(225°)
Harman [31]
1+ \sqrt{3}
tangent 60 is √3 on the unit circle and tangent 225 is 1. So 1+√3 is the exact value of tan 60 plus tan 225
7 0
3 years ago
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2 Points
saw5 [17]
There’s no shape so I think it’s D
3 0
3 years ago
M and N are independent events. P(M | N) = 0.71. Find P(M)
mash [69]

Since M and N are independent events, P(M)=P(M|N)=0.71

Step-by-step explanation:

Two events A and B are said to be independent when the probability of A happening does not depend on whether B has happened or not.

In other word, this condition can be rewrite as follows:

P(A|B)=P(A)

where

P(A|B) is the probability of A happening  given that B has happened

P(A) is the probability of A happening

In this problem, we have:

- M and N are independent events: so M does not depend on whether N occurs or not

Therefore,

P(M)=P(M|N)

- Also, we know that

P(M|N)=0.71

Therefore, it follows that

P(M)=0.71

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6 0
3 years ago
Suppose r(t) = (et cos t)i + (et sin t)j. Show that the angle between r and a never changes. What is the angle?
tatyana61 [14]

Answer:

Angle = Ф = cos^{-1}(0) = 0

Hence, it is proved that angle between position vector r and acceleration vector a = 0 and is it never changes.

Step-by-step explanation:

Given vector r(t) = e^{t}cost i + e^{t}sint j

As we know that,

velocity vector = v = \frac{dr}{dt}

Implies that

velocity vector = (e^{t} cost - e^{t} sint)i + (e^{t} sint - e^{t}cost )j

As acceleration is velocity over time so:

acceleration vector = a = \frac{dv}{dt}

Implies that

vector a =

(e^{t}cost - e^{t}sint - e^{t}sint - e^{t}cost )i + ( e^{t}sint + e^{t}cost + e^{t}cost - e^{t}sint )j

vector a = (-2e^{t}sint) i + ( 2e^{t}cost)j

Now scalar product of position vector r and acceleration vector a:

r. a = . \\

r.a = -2e^{2t}sintcost + 2e^{2t}sintcost

r.a = 0

Now, for angle between position vector r and acceleration vector a is given by:

cosФ = \frac{r.a}{|r|.|a|} = \frac{0}{|r|.|a|} = 0

Ф = cos^{-1}(0) = 0

Hence, it is proved that angle between position vector r and acceleration vector a = 0 and is it never changes.

4 0
3 years ago
Simplify.<br> √20⋅√3·√18<br><br> 5√30<br><br> 5√10<br><br> 6√10<br><br> 6√30
ddd [48]

The correct answer is:   [D]:  " 6√30 " .


_____________________________________


Explanation:


_____________________________________


Simplify:   " √20 ⋅ √3 · √18 " :


___________________________


Step 1)  Simplify the "first term" :

"√20  " = √4 √5 = 2√5  ;


___________________________


Step 2):  Consider the "second term:  

                  "√3 " = √3  (already simplified); 


___________________________


Step 3)  Simplifly the "third term" :

" √18  "  =  √9 * √2 =  " 3√2 " ;


____________________________

We have:  " 2√5 * √3 * 3√2 " ;

                 =   (2* 3) *  (√5 * √3 * √2) ;

                 =  6 * (√5 √3 √2)  ;

Note:  " (√5 * √3 * √2 )"  =   " √(5 * 3 * 2) "  =  " √ 30 " .

So:    " 6 * (√5 * √3 * √2)  =

______________________________________

          →   " 6 √30 " .

______________________________________

The answer is:  " 6√30 " ;

       →  which is:  "Answer choice:  [D]:  " 6√30 " .

_____________________________________

Hope this answer helps!

Best wishes!

_____________________________________

5 0
3 years ago
Read 2 more answers
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