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katrin2010 [14]
3 years ago
15

Which pair of angles is supplementary?

Mathematics
1 answer:
Pie3 years ago
4 0

Answer:

There are 4 possible answers:

1.) Pair 1 and 3

2.) Pair 3 and 4

3.) Pair 4 and 2

4.) Pair 2 and 1

Step-by-step explanation:

Supplementary angles are 2 or more angles who's sum add up to 180°. If two angles make a straight angle they are supplementary.

Hope this helps!

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Pamela is 9 years older than Jiri. The sum of their ages is 95. What is Jiri's age?
Darya [45]
Jiri age = 43
Pamela age = 52

52 - 43 = 9

52 + 43 = 95
4 0
4 years ago
Read 2 more answers
I NEED THIS TO BE CORRECT, I WILL GIVE ANYTHING!!
Andrew [12]

Answer:

a) 105 degrees

b) 75

c) 105

Step-by-step explanation:

a) 180-75 is 105 (supplementary angles)

b) same angle as kings ave and 4th street. (supplementary angles)

c) 180-75 is 105 (also vertical angle rule)

3 0
3 years ago
(8x to the second power plus 17x)+6xto the second power
Phoenix [80]
8x^2+17x+6x^2=14x^2+17x
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3 years ago
Please help with math
Ivan

Answer:

B is the answer

Step-by-step explanation:

8 0
2 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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