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just olya [345]
3 years ago
10

Write the percent as a decimal. 1. 66.7%

Mathematics
2 answers:
Sever21 [200]3 years ago
7 0

Answer:

66.7% = 0.667 in decimal form.

Hope that helps!

juin [17]3 years ago
5 0

Answer:

0.667

Step-by-step explanation:

Percentages can be changed to decimals by dividing by 100 (or moving the decimal left two places).

Decimals can be changed to percentages by multiplying by 100 (or moving the decimal right two places).

Because 66.7% is a percentage (and we're converting to a decimal), we divide by 100 and move the decimal left two places so it becomes 0.667.

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in a football tournament at group stage there are five football teams in a group, Brazil, England, Scotland, Argentina and Franc
emmasim [6.3K]

Answer:

6/25

Step-by-step explanation:

Because england, scotland and france are european team and they make up 3/5 of all the teams, you multiply 3/5 by 2/5 because the south american teams are brazil and argentina which make up 2/5 of the total teams. So the probability that a european team will play a south american team is 3/5*2/5 which is 6/25

7 0
3 years ago
Find each product. <br> (x - 6)(x + 2)
emmainna [20.7K]

Answer:

x^2-4x-12

Step-by-step explanation:

(x-6)(x+2)=x^2-6x+2x-12=x^2-4x-12

6 0
3 years ago
Read 2 more answers
CAN SOMEONE JUST CHECK MY ANSWER?
12345 [234]

Answer:

you are absolutely correct .........

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
Do u agree?I m not pretty sure
nika2105 [10]
It looks correct to me :)
6 0
3 years ago
Read 2 more answers
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