It is known that the three sides of a triangle are: 2, 3 and 4. Is this an obtuse triangle? Why?

It is known that the three sides of a triangle are: 2, 3 and 4. Is this an obtuse triangle? Why?

The length of the opposite side of the maximum angle is 4, and the maximum angle is set as a, so cosa = (2 + 3-4) / 2 × two × 3) = - 1 / 4, so ∠ a ＞ 90 °, so it is an obtuse triangle

Given obtuse triangle trilateral 2,3, x, find the value range of X?

one
. if x is the largest side and is a right triangle,
2*2+3*3＝x^2
X = root 13, so root 13

It is known that a, B and C are the opposite sides of ∠ a, ∠ B and ∠ C in the obtuse angle △ ABC, ∠ C is the obtuse angle, and the area of △ ABC is 5 3, a = 4, B = 5, then C = _

∵ a = 4, B = 5, △ ABC area s = 1
2absinC=5
3，
∴sinC=
three
2，
∵ C is obtuse angle,
∴C=120°，
According to the cosine theorem, C2 = A2 + b2-2abcosc = 16 + 25-20 = 21,
Then c=
21，
So the answer is:
twenty-one

On a straight highway, a car runs at a speed of 15m / s. from a certain moment, it brakes to slow down and move in a straight line. Its acceleration is 2m / s Square. Calculate the vehicle 10 How far is the last car from the braking point in 10s

Push forward from the car stop,
If the vehicle stops after T seconds, then
0+at=v
0+2t=15
T = 7.5 seconds
When the vehicle decelerates, the actual acceleration is - 2m / S ^ 2
Therefore, the driving distance S = VT + 0.5at ^ 2
s=15*7.5-0.5*2*7.5^2
S = 56.25m

The car runs at a constant speed of 30m / s on a straight road. After braking, it decelerates at an acceleration of 5m / s and moves in a straight line. How many meters has the car moved from braking to stopping?

V = V0 + at calculates t = 6, and then calculates the distance from S = v0t + at ^ 2 / 2 to be 90m. The above V represents the final velocity and V0 represents the initial velocity

The car travels at a constant speed of 30 meters per second on a straight road. After braking, it makes a uniform deceleration linear motion at an acceleration of 5 meters per second How many meters did the car go from braking to stopping? 2 from the start of braking timing, the instantaneous speed of the car at the end of the 8th second?

Vt = Vo + at ∵a=-5
When 0 = 30-5t, t = 6
S = Vot + 1/2at ²
S = 30*6 + 1/2*（-5）*6 ² = ninety
∵ the car has stopped at the end of the sixth second, so the instantaneous speed at the end of the eighth second is 0