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Oksi-84 [34.3K]
3 years ago
12

Simplify (3v^3)^2/15v^7

Mathematics
2 answers:
leonid [27]3 years ago
8 0
<span>Combine like terms : (v2 - 5) x (3v + 7)</span>
liraira [26]3 years ago
5 0
   13
3v
------------
     5
(_Please give it the Brainiest answer_)
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Three times the quantity five less than x, divided by the product of six and x
Elden [556K]

Answer:

\frac{3(x-5)}{6x}

Step-by-step explanation:

we know that

The phrase "Three times the quantity five less than x" means 3 multiplied by the difference of x minus 5

3(x-5)

The phrase "the product of six and x" means the number 6 multiplied by x

6x

therefore

The phrase" Three times the quantity five less than x, divided by the product of six and x" is equal to the quotient of 3(x-5) and 6x

\frac{3(x-5)}{6x}

7 0
3 years ago
What is the value of x?
Alisiya [41]
Simplify the first numerator first
-2(x+1)
-2x-2

Then you can set up an equation using cross multiplication
5(-2x-2)= 3x

Simplify and solve
-10x-10=3x
-10=13x
-10/13=x

Final answer: x=-10/13
3 0
3 years ago
What are the solutions to this equation?
ziro4ka [17]

 

\displaystyle\\(x-4)^2=49\\\\(x-4)^2-49=0\\\\(x-4)^2-7^2=0\\\\(x-4-7)(x-4+7)=0\\\\(x-11)(x+3)=0\\\\x-11 = 0~~~\text{or}~~~x+3=0\\\\x-11=0~~~\implies~~~\boxed{x_1=11}\\\\x+3=0~~~\implies~~~\boxed{x_2=-3}\\\\\boxed{\bf The~solutions~are:~~x =11~~\text{or}~~x =-3}



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
The width of a rectangle is 4x-4 feet and the length is 2x+5 feet. Find the perimeter
Ipatiy [6.2K]
Perimeter is
2 • width+ 2•length=
2• (4x-4)+2•(2x+5)=
Distribute multiplication in parentheses
8x-8+4x+10=
Combine like terms
12x+2
4 0
3 years ago
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