Answer:
The value of the side PS is 26 approx.
Step-by-step explanation:
In this question we have two right triangles. Triangle PQR and Triangle PQS.
Where S is some point on the line segment QR.
Given:
PR = 20
SR = 11
QS = 5
We know that QR = QS + SR
QR = 11 + 5
QR = 16
Now triangle PQR has one unknown side PQ which in its base.
Finding PQ:
Using Pythagoras theorem for the right angled triangle PQR.
PR² = PQ² + QR²
PQ = √(PR² - QR²)
PQ = √(20²+16²)
PQ = √656
PQ = 4√41
Now for right angled triangle PQS, PS is unknown which is actually the hypotenuse of the right angled triangle.
Finding PS:
Using Pythagoras theorem, we have:
PS² = PQ² + QS²
PS² = 656 + 25
PS² = 681
PS = 26.09
PS = 26
Answer:
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Step-by-step explanation:
Answer:
The mentioned number in the exercise is:
Step-by-step explanation:
To obtain the mentioned number in the exercise, first you must write the equations you can obtain with it.
If:
- x = hundredths digit
- y = tens digit
- z = ones digit
We can write:
- x = z + 1 (the hundreds digit is one more than the ones digit).
- y = 2x (the tens digit is twice the hundreds digit).
- x + y + z = 11 (the sum of the digits is 11).
Taking into account these data, we can use the third equation and replace it to obtain the number and the value of each digit:
- x + y + z = 11
- (z + 1) + y + z = 11 (remember x = z + 1)
- z + 1 + y + z = 11
- z + z +y + 1 = 11 (we just ordered the equation)
- 2z + y + 1 = 11 (z + z = 2z)
- 2z + y = 11 - 1 (we passed the +1 to the other side of the equality to subtract)
- 2z + y = 10
- 2z + (2x) = 10 (remember y = 2x)
- 2z + 2x = 10
- 2z + 2(z + 1) = 10 (x = z + 1 again)
- 2z + 2z + 2 = 10
- 4z + 2 = 10
- 4z = 10 - 2
- 4z = 8
- z = 8/4
- <u>z = 2</u>
Now, we know z (the ones digit) is 2, we can use the first equation to obtain the value of x:
- x = z + 1
- x = 2 + 1
- <u>x = 3</u>
And we'll use the second equation to obtain the value of y (the tens digit):
- y = 2x
- y = 2(3)
- <u>y = 6</u>
Organizing the digits, we obtain the number:
- Number = xyz
- <u>Number = 362</u>
As you can see, <em><u>the obtained number is 362</u></em>.
Answer:
x = 28.01t,
y = 10.26t - 4.9t^2 + 2
Step-by-step explanation:
If we are given that an object is thrown with an initial velocity of say, v1 m / s at a height of h meters, at an angle of theta ( θ ), these parametric equations would be in the following format -
x = ( 30 cos 20° )( time ),
y = - 4.9t^2 + ( 30 cos 20° )( time ) + 2
To determine " ( 30 cos 20° )( time ) " you would do the following calculations -
( x = 30 * 0.93... = ( About ) 28.01t
This represents our horizontal distance, respectively the vertical distance should be the following -
y = 30 * 0.34 - 4.9t^2,
( y = ( About ) 10.26t - 4.9t^2 + 2
In other words, our solution should be,
x = 28.01t,
y = 10.26t - 4.9t^2 + 2
<u><em>These are are parametric equations</em></u>
