Answer:
5.266 secs
Step-by-step explanation:
Lets assume ; p(t) = t^-3 + 2^2 + ( 3/2 ) is the particle position along x-axis
time interval [ 0, 4 ]
Average velocity = Displacement / time
= p( b ) - p( a ) / b - a -------- ( 1 )
where a = 0 , b = 4 ( time intervals )
Back to equation 1
Average velocity = [ ( 4^-3 + 4 + (3/2) ) - ( 0 + 4 + (3/2) ) ] / 4
= 3.9 * 10^-3 ----- ( 2 )
Instantaneous velocity = d/dx p(t)
= - 3/t^4 ------ ( 3 )
To determine the time that the instanteous velocity = average velocity
equate equations (2) and (3)
3.9*10^-3 = - 3 / t^4
t^4 = - 3 / ( 3.9 * 10^-3 ) = - 769.231
hence t = ![\sqrt[4]{- 769.231 }](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B-%20769.231%20%7D)
= 5.266 secs
we ignore the negative sign because time can not be in the negative
Answer:
1) ∠A=84°
2) ∠C=20°
Step-by-step explanation:
1)
First, find ∠C:
<em>(I'm assuming the exterior angle of 126° makes a straight line with ∠C)</em>
The angles on a straight line always add up to 180. Therefore:
∠C+126=180
∠C=180-126
∠C=54
Then find ∠B:
We also know that all the angles in a triangle add up to 180. Therefore:
∠A+∠B+∠C=180
∠A+∠B+54=180
∠A+∠B=126
<em>(we know ∠A=2(∠B))</em>
2(∠B)+∠B=126
3(∠B)=126
∠B=42
Now, find ∠A:
∠A=2(∠B)
∠A=2(42)
∠A=84°
2)
First, find ∠B:
<em>(Again, I'm assuming the exterior angle of 100° makes a straight line with ∠B)</em>
The angles on a straight line always add up to 180. Therefore:
∠B+100=180
∠B=180-100
∠B=80
Then find ∠A:
We also know that all the angles in a triangle add up to 180. Therefore:
∠A+∠B+∠C=180
∠A+80+∠C=180
∠A+∠C=100
<em>(we know ∠A=4(∠C))</em>
4(∠C)+∠C=100
5(∠C)=100
∠C=20°
Let's start b writing down coordinates of all points:
A(0,0,0)
B(0,5,0)
C(3,5,0)
D(3,0,0)
E(3,0,4)
F(0,0,4)
G(0,5,4)
H(3,5,4)
a.) When we reflect over xz plane x and z coordinates stay same, y coordinate changes to same numerical value but opposite sign. Moving front-back is moving over x-axis, moving left-right is moving over y-axis, moving up-down is moving over z-axis.
A(0,0,0)
Reflecting
A(0,0,0)
B(0,5,0)
Reflecting
B(0,-5,0)
C(3,5,0)
Reflecting
C(3,-5,0)
D(3,0,0)
Reflecting
D(3,0,0)
b.)
A(0,0,0)
Moving
A(-2,-3,1)
B(0,-5,0)
Moving
B(-2,-8,1)
C(3,-5,0)
Moving
C(1,-8,1)
D(3,0,0)
Moving
D(1,-3,1)
The given operation involves just swapping the two rows, so carrying out
on the matrix gives
![\left[\begin{array}{cc|c}5&5&5\\1&-\dfrac23&5\end{array}\right]\to\left[\begin{array}{cc|c}1&-\dfrac23&5\\5&5&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Cc%7D5%265%265%5C%5C1%26-%5Cdfrac23%265%5Cend%7Barray%7D%5Cright%5D%5Cto%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Cc%7D1%26-%5Cdfrac23%265%5C%5C5%265%265%5Cend%7Barray%7D%5Cright%5D)
