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Masja [62]
3 years ago
15

Find the measure of z. A. 80° B. 83 ° C. 70 ° D. 87 °

Mathematics
1 answer:
Ymorist [56]3 years ago
7 0
The answer to the question is c. 70°
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Several members at a gym were asked how many minutes they worked out at the gym that day as well as how many calories they burne
n200080 [17]

Answer:

your welcome

Step-by-step explanation:

he said thank you

4 0
3 years ago
The interior angles of a triangle have measures k°, 27°, and 10°. What is the value of k?ANSWER MY FREAKING QUESTION
lapo4ka [179]
143o

interior angles of a triangle always add up to 180o
6 0
3 years ago
Find two consecutive numbers whose squares differ by 45
Wittaler [7]

Answer:

The 2 numbers are 22 and 23

Step-by-step explanation:

let the two numbers be x and x+1

according to the question:

(x+1)^2 - x^2 = 45

x^2+1+2x - x^2 = 45

2x + 1 = 45

2x= 44

x = 22

the other number is x+1 = 22+1 = 23

therefore, the two consecutive numbers are 22 and 23

Hope it helped u....

5 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
May I get some help on this please?
HACTEHA [7]

Answer:

2

Step-by-step explanation:

The answer is the second explanation I think

8 0
3 years ago
Read 2 more answers
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