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3 years ago

Hellppp 70 pointsss

Mathematics

1 answer:

dimulka [17.4K]3 years ago

7 0

y=a.(X-x1)(X-x2)

x1 and x2 are x-intercepts

x1=-3

x2=1

other points: (-1,8) and (0,5)

y=a.(x+3).(x-1)

a=-2

f(x)= -2.(x+3)(x-1)

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Round to the nearest tenth.

laila [671]

Answer:

8.2 units

Step-by-step explanation:

Given that the only information for our triangle are 2 sides and 1 angle, we must use the Law of Cosines to find side BC

Recall the Law of Cosines

$a^2=b^2+c^2-2bc\cdot cos(A)$

Identify angles and sides

$m\angle A=23^\circ\\a=BC=?\\b=AC=14\\c=AB=6$

Solve for side "a"

$a^2=b^2+c^2-2bc\cdot cos(A)\\\\a^2=14^2+6^2-2(14)(6)\cdot cos(23^\circ)\\\\a^2=196+36-168cos(23^\circ)\\\\a^2=222-168cos(23^\circ)\\\\a=\sqrt{222-168cos(23^\circ)}\\ \\a\approx8.2$

Therefore, the length of line segment BC is about 8.2 units

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2 years ago

For the amusement of the guests, some hotels have elevators on the outside of the building. One such hotel is 400 feet high. You

Mariana [72]

9514 1404 393

Answer:

a. h(t) = -20t +400

b. θ(t) = arctan(2 -2/15t); dθ/dt = -30/(1125 -120t +4t^2)

c. 15 seconds; 100 ft

Step-by-step explanation:

a. The initial height of the elevator is 400 ft. The rate of change of height is -20 ft/s, so the height equation can be ...

h(t) = -20t +400

b. The tangent of the angle above the line of sight is "opposite"/"adjacent":

tan(θ) = (h(t) -100)/(150) = -2/15t +2

θ(t) = arctan(2 -2/15t) . . . . radians

The derivative of the angle function is ...

dθ/dt = 1/(1+(2 -2/15t)^2)(-2/15)

dθ/dt = -30/(1125 -120t +4t^2)

c. The value of dθ/dt will have a peak where the denominator has a minimum, at t = -(-120)/2(4)) = 15. (The quadratic vertex coordinate is t=-b/(2a).)

The elevator appears to be moving fastest at t=15 seconds.

The height at that time is ...

h(15) = 400 -20(15) = 100

The elevator appears to be moving fastest when it is at eye level, 100 ft above the ground.

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3 years ago

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xeze [42]

I believe your answer would have to be 12

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