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Advocard [28]
3 years ago
5

5

Mathematics
1 answer:
dalvyx [7]3 years ago
6 0

Answer: No solution

Step-by-step explanation:

1. Expand the brackets- 9-3x+5x=2x+4

2. Add common terms- 9+2x=2x+4

3. Subtract 2x from both sides- 9=4

4. Nine cannot equal four and therefore, there is no solution.

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Find the slope between the points (-8,-4) and (0,0).
Vadim26 [7]

Answer:

1/2

Step-by-step explanation:

slope = \frac{y_{2}-y_{1} }{x_{2}-x_{1}}

(0-(-4))/(0-(-8)) = 1/2

7 0
3 years ago
HURRY! What is the value of x in the diagram? Please leave an explanation fr the answer
melomori [17]
X = -15 because first u move the ax’s to one side so it would be 3x-5x=25+5 then u calculate each side 3x-5x= -2x and 25+5=30 so -2x=30 and then u divide both sides by -2 and -2/-2 cancels out and 30/-2 equals -15 (I hope his helps) :))
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3 years ago
The original price of a plane ticket was reduced by $150.
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3 years ago
Read 2 more answers
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
Solve for x. <br><br> a²x+(a-8)=(a+8)x
Sergio [31]
a^2x+(a-8)=(a+8)x\\\\(a+8)x=a^2x+(a-8)\ \ \ \ |subtract\ a^2x\ from\ both\ sides\\\\(a+8)x-a^2x=a-8\\\\(a+8-a^2)x=a-8\ \ \ \ |divide\ both\ sides\ by\ (a+8-a^2)\\\\\huge\boxed{x=\frac{a-8}{a+8-a^2}}\to\huge\boxed{x=\frac{8-x}{a^2-a-8}}
6 0
3 years ago
Read 2 more answers
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