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r-ruslan [8.4K]
3 years ago
13

Craig like to collect records last year he had 12 records in his collection now he has 15 records what is the percent increase o

f his collection
Mathematics
2 answers:
Sergio039 [100]3 years ago
7 0
It should be a 20% increase
erastovalidia [21]3 years ago
3 0
25% increase 1/4 of 12 1/4 is equal to 25%
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martha bought an apple for $0.89 and a drink for $1.95 which is the best estimate of how much money she spent? can someone plz h
77julia77 [94]
Okay first we add them:
0.89 + 1.95 = 2.84
Then we estimate them to the tenths place:
2.84 (estimated to) 2.80
Since the number in the ones place is less than 5, we estimate to 80.
6 0
3 years ago
Read 3 more answers
0.333... as a fraction it’s repeating fraction and 0.666 as a repeating fraction
andre [41]

Answer:

0.333 is 1/3

0.666 is 2/3

Step-by-step explanation:

I used my head but you can use a calculator too.

7 0
3 years ago
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How do i get the answer x= 3, -1 for equation (3/x^2)+ 2/x =1
Assoli18 [71]

Answer:

Steps below

Step-by-step explanation:

(3/x²)+ 2/x =1   (x ≠ 0)

(3/x²) + 2x/x² = 1

(3 + 2x) / x² = 1

3 + 2x = x²

x² - 2x - 3 = 0

(x + 1) (x - 3) =0

x = -1 or x = 3

3 0
3 years ago
The number 12.5 can be classified as:
agasfer [191]

Answer:

ur gay

Step-by-step explanation:

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
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