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sineoko [7]
3 years ago
15

Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.

Mathematics
2 answers:
Rainbow [258]3 years ago
7 0

Answer:

z=108

Step-by-step explanation:

z÷12=9

x12 x12

z=108

vredina [299]3 years ago
3 0
Answer: z= 108

Explanation:

You might be interested in
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
Lisa bought an outfit for<br> $37.67. What is $37.67<br> rounded to the nearest<br> whole number?
almond37 [142]

Answer:

Answer is 38

Step-by-step explanation:

if the first digit in the fractional part of 37.67 is less than 5 then you remove the fractional part to get the answer. If the first digit is 5 or above, then you add 1 to the integer and remove the fractional part to get the answer. First digit is 6 - it is 5 and above, you add 1 to the integer and remove fractional part and get 37.67 rounded to the nearest whole number as 38

Hope this helps

8 0
2 years ago
Read 2 more answers
‼️helpp ASAP pleasee
koban [17]
We know the measure of one of the angles.Angle b which is 90.

Now we have to find the other two angles.

First we will find angle a.
There are 180 degrees in a straight line.
Sooo...
180-130=50
Angle a will be 50.

There are also 180 degrees in a triangle.
To find the last angle we do this
180-50-71=59

Your answer and the measure of angle c will be 59 or option 3. 

7 0
3 years ago
approximate each irrational number to the nearest hundredth without using a calculator square root of 118 and 319​
Strike441 [17]

Answer:

\sqrt{118}\approx 10.86

\sqrt{319}\approx 17.86

Step-by-step explanation:

Consider the provided number.

We need to find the approximate value of \sqrt{118} to the nearest hundredth.

First find two perfect squares that the irrational number falls between.

100

118 is lying between 100 and 121, therefore the square root value of 118 will be somewhere between 10 and 11.

\sqrt{100}

10

118 is closer to 121 as compare to 100.

Therefore, \sqrt{118}\approx 10.86

Consider the number \sqrt{319}

First find two perfect squares that the irrational number falls between.

289

319 is lying between 289 and 324, therefore the square root value of 319 will be somewhere between 17 and 18.

\sqrt{289}

17

319 is closer to 324 as compare to 289.

Therefore, \sqrt{319}\approx 17.86

8 0
3 years ago
sketch a unit circle with angle instandard positionquestions:a. For what values of 0 is the sine increasing? Decreasing?b. For w
liq [111]

We can draw a unit circle as:

The sine is the vertical leg (or vertical projection) of the triangle formed by the angle.

The cosine will be the horizontal leg (or horizontal projection) of this same triangle.

The hypotenuse of this triangle will be the radius if the unit circle (r = 1).

a) The sine will be increasing from θ = 0° until it reaches its maximum at 90°.

Then it will be decreasing for angles between 90° and 270°.

Between 270° and 0° it will be increasing again.

b) The cosine has a maximum value for an angle of 0°.

It will decrease until 180°.

Then it will be increasing from angles between 180° and 360°.

c) We can see in the unit circle that the sine is 0 when the angle is one the horizontal axis, like when θ = 0° or θ = 180°.

The cosine will be equal to 0 when the angle is vertical: θ = 90° and θ = 270°.

Answer:

a) Increasing: (0°, 90°) and (270°, 360°)

Decreasing: (90°, 270°)

b) Increasing: (180°, 360°)

Decreasing: (0°,180°).

c) Sin(θ) = 0 for θ = 0° and θ = 180°

Cos(θ) = 0 for θ = 90° and θ = 270°

4 0
1 year ago
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