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meriva
4 years ago
9

Multiply (x-2)(3x+4) using the foil method

Mathematics
1 answer:
masya89 [10]4 years ago
4 0
<span>Multiply (x-2)(3x+4) using the foil method 

</span>(x-2)(3x+4) =  \\  \\ (x*3x) +(x*4) +(-2*3x) +(-2*4) =  \\  \\ 3x^2 + 4x - 6x - 8 =  \\  \\ 3x^2 -2x - 8  \\  \\  \\  \boxed{(x-2)(3x+4) =3x^2 -2x - 8}<span>


</span>
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One student from the school will be selected at random. What is the probability the selected student chose the art elective and
Black_prince [1.1K]

Answer:

0

Step-by-step explanation:

The students are only allowed to take one slot for the elective. Therefore, there is a probability of zero that they would have chosen both art and music.

4 0
3 years ago
Can someone do these 20 problems for me I’ll make whoever answers first the Brainlylist … Due today
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(31x15)108divided by6
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8370

Step-by-step explanation:

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3 years ago
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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
4 years ago
1. A lab has two bacteria cultures. Culture A
GarryVolchara [31]

Answer:

A.

Step-by-step explanation:

To get the answer, you will need to divide the number of bacterias that Culture A has by the number of bacterias that Culture B has.

Size=(8*104)/(4*104)

=2

So, A will be the correct answer. (I have corrected your answer's no. of bacterias that Culture B contains as 104.)

7 0
3 years ago
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