All categories

3 years ago

Please help The question is on Mathamatics

Mathematics

1 answer:

sweet [91]3 years ago

4 0

Answer:

A
(The probability of randomly selecting a picture that shows Justin with his friends is greater than the probability of randomly selecting a picture that shows Justin with his family.)

You might be interested in

Whats 100,002÷13 dont try calculator or paper

balandron [24]

100,002 divded by 13 will be 7692 when estimated

6 0

3 years ago

Read 2 more answers

How do you find DE, Round to nearest 10th

Sergeu [11.5K]

Answer: $DE\approx6.0$

Step-by-step explanation:

Observe in the figure given in the exercise that four right triangles are formed.

In this case you can use the following Trigonometric Identity to solve this exercise:

$cos\alpha=\frac{adjacent}{hypotenuse}$

From the figure you can identify that:

$\alpha =53\°\\\\hypotenuse=10\\\\adjacent=BE=DE$

Then, you can substitute values:

$cos(53\°)=\frac{DE}{10}$

The next step is to solve for DE in order to find its value. This is:

$10*cos(53\°)=DE\\\\DE=6.01$

Finally, rounding the result to the nearest tenth, you get that this is:

$DE\approx6.0$

7 0

3 years ago

What is 37 and 1/2 simplified?

yarga [219]

37 and 1/2 could mean:
37+ (1/2)
37 times 1/2

37+1/2=37 and 1/2 or 37.5

37 times 1/2=37/2=18 and 1/2=18.5

3 0

4 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

What is the value of 3 in 384?<br> Please explain?!

seropon [69]

Answer:uhhhhh 300

Step-by-step explanation:

I don’t really need to explain bc all u need to do is take the 80 and 4 away and u have only 300

3 0

3 years ago