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pochemuha
3 years ago
15

Which statement is true?

Mathematics
1 answer:
adell [148]3 years ago
4 0

Answer:

Answer is the 2nd one.

Step-by-step explanation:

You might be interested in
One number is four times another number. If their sum is 95, what are the numbers?
nikdorinn [45]
Let x be one number and y be the other.

#1 - Set up a system of equations.
4x = y
x + y = 45

#2 - Solve the system by substitution.
x + (4x) = 45
5x = 45
x = 9

(9) + y = 45
y = 36
8 0
3 years ago
Read 2 more answers
Compare the theoretical probability to the experimental probability of landing on H.
cestrela7 [59]

The correct statement comparing the theoretical and experimental probabilities is given as follows:

\frac{1}{4} < \frac{7}{25}.

<h3>What is a probability?</h3>

A probability is given by the <u>number of desired outcomes divided by the number of total outcomes</u>.

The theoretical probability is taken before any experiment. Since the four sections are equal, the theoretical probability is:

T(H) = 1/4.

The experimental probability is taken considering previous experiments. Out of 100 tosses, 28 landed on H, hence:

E(H) = 28/100 = 7/25.

Hence the correct statement is:

\frac{1}{4} < \frac{7}{25}.

More can be learned about probabilities at brainly.com/question/14398287

#SPJ1

4 0
2 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
Answer please.....<br> Thanks!!!!
Tems11 [23]
A. x= 11/5
b. x= 6
c. x= -12/2
d. x= 6/5
3 0
3 years ago
What is the domain and range for the following function and its inverse?
Basile [38]

Answer:

"f(x)  

domain: all real numbers, range: all real numbers  

f–1(x)  

domain: all real numbers, range: all real numbers"

Step-by-step explanation:

We can use the fact that the domain of a function and the range of its inverse are equal.

Also, the range of the function and the domain of its inverse are equal as well.

<em>Looking at the function f(x/ = -x + 5, we see that this is a line with a negative slope of 1 and a y-intercept of +5. </em>

As we know from the graph of lines, there is no restricting values in x and y. So for the original function,  domain is the set of all real numbers and the range is the set of all real numbers.

For the inverse, the range is set of all real numbers and domain is also the set of all real numbers.

First answer choice is right.

4 0
3 years ago
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