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3 years ago

What is 1 1/2 times 3?

Mathematics

2 answers:

zepelin [54]3 years ago

7 0

4 1/2 is 1 1/2 times 3

Katarina [22]3 years ago

4 0

$1 \frac{1}{2} = \frac{(1\cdot2) + 1}{2} = \frac{3}{2}$

$\frac{3}{2} \cdot 3 =\boxed{ \boxed{\boxed{\boxed{\boxed{\boxed{\bf{\frac{9}{2} = 4.5}}}}}}}}}}$

So your answer as a fraction is 9/2, while your answer as a decimal is 4.5.

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Four to the sixth power times four to the -8th power please explain how to do this I don't understand it

Alenkinab [10]

Your question equals 1/16 or 0.0625.

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3 years ago

Which function is shown in the graph below?

Helga [31]

Answer:

Step-by-step explanation:

I’m not entirely sure if this is right but I hope I helped.

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3 years ago

Claire is putting up a fence around her rectangular garden. The length is four less than double the width. The perimeter is 22 f

Doss [256]

Answer:

The dimensions are 6 by 5 feet.

Length = 6 feet.

Width = 5 feet.

Step-by-step explanation:

Let the length = L

Let the width = W

Perimeter of a rectangle = 2L + 2W

Translating the word problem into an algebraic equation, we have;

L = 2W - 4 ........equation 1

22 = 2L + 2W .......equation 2

Substituting the value of "L" into equation 2, we have;

22 = 2(2W - 4) + 2W

22 = 4W - 8 + 2W

22 + 8 = 6W

30 = 6W

W = 30/6

Width, W = 5 feet.

To find the length, L

Substituting the value of "W" into equation 1, we have;

L = 2W - 4

L = 2(5) - 4

L = 10-4

Length, L =6 feet

Therefore, the dimensions of the garden are 6 by 5 feet.

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3 years ago

2. Kevin asked Olivia what parallel lines are. Olivia responded, "They are lines that never

Shkiper50 [21]

A I hope this really helps

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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