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Katyanochek1 [597]
3 years ago
13

What is another why to say geometry​

Mathematics
2 answers:
nasty-shy [4]3 years ago
8 0
Configuration. Hope this helps! :)
lord [1]3 years ago
6 0

Answer:

jeeoumetry

Step-by-step explanation:

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What’s the scale factor
dem82 [27]

Answer:

Step-by-step explanation:

The scale is depending on the size of each cat. Lets say the larger cat is A and the smaller one is B. The scale factor from B to A is 1.75 (or 3.5/2)

when solving for scale factor you divide the original drawing and scale drawing. Original is denominator, Scale is Numerator.

4 0
3 years ago
Caylan is making bakes goose for a charity bake sale ha places a tray of scones a tray of muffing and a tray of cookies in the o
Varvara68 [4.7K]

Answer:

Step-by-step explanation:

an hour, or 60 minutes

3 0
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
How many counters would you place in the five frame to show the number
Leviafan [203]
Assuming the five frame is for the number five (5) then place 5 counters in the frame.
7 0
3 years ago
An exponential function f(x) = ab^xpasses through the points (0, 2) and (3, 54). What are the values of a and
Vsevolod [243]

Answers:

a = 2

b = 3

=======================================================

Explanation:

Plug in x = 0 and y = 2 to find that

y = a*b^x

2 = a*b^0

2 = a*1

2 = a

a = 2

Then plug in x = 3 and y = 54 to determine the value of b

y = a*b^x

y = 2*b^x

54 = 2*b^3

2b^3 = 54

b^3 = 54/2

b^3 = 27

b = (27)^(1/3)

b = 3

So we have y = a*b^x update to y = 2*3^x

7 0
3 years ago
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