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Gelneren [198K]
3 years ago
9

Consider △LNM. Which statements are true for triangle LNM? Check all that apply. The side opposite ∠L is NM. The side opposite ∠

N is ML. The hypotenuse is NM. The hypotenuse is LN. The side adjacent ∠L is NM. The side adjacent ∠N is ML.
Mathematics
2 answers:
Anvisha [2.4K]3 years ago
6 0

The following statements are true by definition:

The side opposite ∠L is NM.

The side opposite ∠N is ML.

The side opposite to the angle should not contain any letter of that side.

 

<span>The following statements are not essentially true because we have no idea if triangle LNM is a right triangle (if it is, then we do not know what the hypotenuse is):</span>

The hypotenuse is NM.

The hypotenuse is LN.

 

<span>The following statements are not true:</span>

The side adjacent ∠L is NM.

The side adjacent ∠N is ML.

They are not true because the side adjacent to an angle should have its letter on the side. For example, the side adjacent to ∠L should be LN or LM and for ∠N it should be NM or NL.

<span> </span>

PolarNik [594]3 years ago
6 0

Answer:

A,1.The side opposite ∠L is NM.

B,2.The side opposite ∠N is ML.

D,4.The hypotenuse is LN.

Those are the correct answers.


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Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'
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Answer:

a) v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9

a(t) = x''(t) = v'(t) =6t-12

b)  0

c) a(t) = x''(t) = v'(t) =6t-12

When the acceleration is 0 we have:

6t-12=0, t =2

And if we replace t=2 in the velocity function we got:

v(t) = 3(2)^2 -12(2) +9=-3

Step-by-step explanation:

For this case we have defined the following function for the position of the particle:

x(t) = t^3 -6t^2 +9t -5 , 0\leq t\leq 10

Part a

From definition we know that the velocity is the first derivate of the position respect to time and the accelerations is the second derivate of the position respect the time so we have this:

v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9

a(t) = x''(t) = v'(t) =6t-12

Part b

For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:

v(t) = 3t^2 -12t +9

We can factorize this function as v(t)= 3 (t^2- 4t +3)=3(t-3)(t-1)

So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is 0\leq t\leq 10 we need to analyze the following intervals:

0< t

For this case if we select two values let's say 0.25 and 0.5 we see that

v(0.25) =6.1875, v(0.5)=3.75

And we see that for a=0.5 >0.25=b we have that f(b)>f(a) so then the function is decreasing on this case.  

1

We have a minimum at t=2 since at this value w ehave the vertex of the parabola :

v_x =-\frac{b}{2a}= -\frac{-12}{2*3}= -2

And at t=-2 v(2) = -3 that represent the minimum for this function, we see that if we select two values let's say 1.5 and 1.75

v(1.75) =-2.8125< -2.25= v(1.5) so then the function sis decreasing on the interval 1<t<2

2

We see that the function would be increasing.

3

For this interval we will see that for any two points a,b with a>b we have f(a)>f(b) for example let's say a=3 and b =4

f(a=3) =0 , f(b=4) =9 , f(b)>f(a)

The particle is moving to the right then the velocity is positive so then the answer for this case is: 0

Part c

a(t) = x''(t) = v'(t) =6t-12

When the acceleration is 0 we have:

6t-12=0, t =2

And if we replace t=2 in the velocity function we got:

v(t) = 3(2)^2 -12(2) +9=-3

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Answer:

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Sum of all 20 student's age = 264

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Sum of current 18 student's age = 234

New avaerage if 2 students (15 year old) left = 234 / 18

New avaerage if 2 students (15 year old) left = 13

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