<span>We have a right angled triangle with an opposite of 300.5 ft. (306 - 5.5) and an adjacent of 400 ft. Recalling SOH CAH TOA, tanθ = O/A.
tan(θ) = 300.5/400.
θ = tan^-1(300.5/400).
θ = 36.9°.</span>
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Answer:</h3>
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Step-by-step explanation:</h3>
If the trapezoid is isosceles, angles A and C are supplementary, so ...
... (4x+4) +(7x+11) = 180
... 11x +15 = 180 . . . . . . collect terms
... 11x = 165 . . . . . . . . . subtract 15
... x = 15 . . . . . . . . . . . . divide by the coefficient of x
And angles C and E are congruent.
... 4·15 +4 = 21y +1
... 63 = 21y . . . . . subtract 1
... 3 = y . . . . . . . . divide by the coefficient of y
Answer:
The measure of other three angles are 135° , 45° and 45°
Step-by-step explanation:
The opposite angles of a parallelogram are equal.
So if we consider a parallelogram ABCD ,
Angle A = Angle C
Angle B = Angle D
If we take angle A = 135° then angle C = 135°
SUM OF INTERIOR ANGLES OF A QUADRILATERAL IS 360°.
A+B+C+D=360°
135°+B+135°+D=360°
270°+B+D=360°
But B=D
So,2B=90°
B=45°
D=45°
So A = C = 135°
A = C = 135° B = D = 45°
A = C = 135° B = D = 45°
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Answer:
Step-by-step explanation:
The extrema will be at the ends of the interval or at a critical point within the interval.
The derivative of the function is ...
f'(x) = 3x² -4x -4 = (x -2)(3x +2)
It is zero at x=-2/3 and at x=2. Only the latter critical point is in the interval. Since the leading coefficient of this cubic is positive, the right-most critical point is a local minimum. The coordinates of interest in this interval are ...
f(0) = 2
f(2) = ((2 -2)(2) -4)(2) +2 = -8 +2 = -6
f(3) = ((3 -2)(3) -4)(3) +2 = -3 +2 = -1
The absolute maximum on the interval is f(0) = 2.
The absolute minimum on the interval is f(2) = -6.
