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sweet [91]
3 years ago
12

What is the value of x? Round the answer to the nearest tenth

Mathematics
1 answer:
arsen [322]3 years ago
7 0

Answer:

The answer is A.

Step-by-step explanation:

The answer is A because an obtuse angle rounds to 120 degrees, then you divide it by 6 which is A.

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What is 14/74 in simplest form
charle [14.2K]
14/74 in the simplest form is 7/37
5 0
3 years ago
A car used 15 gallons of gasoline and traveled a total distance of 350 miles. The car's fuel efficiency is 25 miles per gallon o
stepan [7]
The equation would be letter b. 25h + 20 (15 - h) = 350
Let’s try to solve to get the value of h
=> 25h + 20(15 -h) = 350
=> 25h + (300 – 20h) = 350
=> 25h + 300 – 20h = 350
=> 300 + 5h = 350
=> 5h = 350 – 300
=> 5h / 5 = 50/ 5
=> h = 10
Thus the value of h is 10 which means a car travel on the highways and used 10 gallons of gasoline,



3 0
3 years ago
Can someone please help me on this question?
Olin [163]
The ends of the graph will extend in opposite directions.

Hope this helped!
6 0
3 years ago
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'
vodka [1.7K]

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

5 0
3 years ago
Write the expression using exponents.<br> 2.1•x•z•z•z•z
Sveta_85 [38]

Answer:

I believe it's 2.1xz^4

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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