We will calculate it like this, We will make a right angle triangle by adding one new point called O for example, and O is located right under the point C, and on the right side of point D, so the coordinated for this point are O (8, -2). From there, length CD is calculated using the Pythagorean theorem, because we know that the CO = 4 and DO = 4, so:
CD^2 = CO^2 + DO^2
CD^2 = 4^2 + 4^2
DC^2 = 16 + 16
DC^2 = 32
DC = square root of 32 = square root of 16 * square root of 2
DC = 4 * square root of 2
The correct answer is B. 4*square root of 2.
Answer: f(x)/g(x) = 3x^2 + 6x - 2 - 12/3x+1
Step-by-step explanation:
The first function is f(x) = 9x^3+21x^2-14
The second function is g(x) = 3x+1
f(x)/g(x = 9x^3+21x^2-14/3x+1
We perform the long division as shown in the attachment to obtain the quotient as: Q(x) = 3x^2+6x - 2 and remainder R = -12
Therefore f(x)/g(x) = 3x^2 + 6x - 2 - 12/3x-1
Where x does not = -1/3
take a factor of 6 out to make 6(4x-3)
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
