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Reil [10]
4 years ago
13

2/3x+1/2=1/6. I need help so help me please

Mathematics
2 answers:
earnstyle [38]4 years ago
7 0

Answer: -1/2

Step-by-step explanation: First, multiply both sides by 6. Move the constant to the right. Calculate and then Divide both sides by 4.

lara [203]4 years ago
6 0

Answer: x=-1/2

Step-by-step explanation:

2/3x+1/2=1/6

Subtract the 1/2

2/3x=-1/3

Divide by 2/3

x=-1/2

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The some of the numbers is 77. The greater number is 5 more than the smaller number.Which equation can be used to solve for the
QveST [7]

Let the smaller number be x and the larger one be y

x + y = 77

y = x + 5

Substituting into the first equation

x + (x + 5) = 77

2x + 5 = 77

Subtracting 5 to both sides

2x +5 -5 = 77-5

2x = 72

Divide both sides by 2

2x/2 = 72/2

x = 36

8 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
Simplify (8^4)−^16<br> Please and thank you
____ [38]

Answer:

(2³)⁴- (2⁴)

=(4096)-(16)

4018

3 0
3 years ago
Particle x moves along the positive x-axis so that it's position at time t is less than equal to 0 is given by x(t)= 5t^3 - 9t^2
serg [7]

Answer:

<h3>YES</h3>

Step-by-step explanation:

Given the position of the particle expressed using the expression;

x(t)= 5t^3 - 9t^2 + 7 where;

x moves along the positive x-axis

when t = 1

x(1) = 5(1)^3 - 9(1)^2 + 7

x(1) = 5 - 9 + 7

x(1) = -4 + 7

x(1) = 3

<em>Since the position of the particle is positive at t = 1, hence the particle is moving towards the right at time t = 1</em>

8 0
3 years ago
Which graph shows a dilation?
galina1969 [7]

Answer:

here you go

Step-by-step explanation:

yes graph is dilation

3 0
3 years ago
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