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2 years ago

Construct an inscribed circle in triangle PQR by finding the incenter of the triangle.

Mathematics

1 answer:

zhannawk [14.2K]2 years ago

6 0

Answer:

i dont know SORRY

Step-by-step explanation:

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(-2.3) + (-5.9) Can somebody please show me how to do these types of problems?

agasfer [191]

Answer:

Idk

Step-by-step explanation: I think u multiply the -2.3 and -5.9 and get 13.57. Idk about the addition part

7 0

3 years ago

Complete the statement of congruence for the triangles.

Hoochie [10]

If you are proving a congruence of triangles then you need to match the corresponding parts to the triangle you are proving congruent.

BAC would then be congruent to MJR because of the corresponding angles

B is the same angle as M

A is the same angle as J

C is the same angle as R

3 0

3 years ago

The function<img src="https://tex.z-dn.net/?f=f%28x%29%3D-3x%5E%7B3%7D%20%2Bx%5E%7B2%7D%20%2B2x" id="TexFormula1" title="f(x)=-3

musickatia [10]

Answer:

True

Step-by-step explanation:

True.

Like the cubic term -3x^3 is negative, for small values of "x", the dependent variable "y" will rise. All this can be verified by looking at the graph.

6 0

3 years ago

A car traveling east at 40.0 m/s passes a trooper hiding at the roadside. The driver uniformly reduces his speed to 25.0 m/s in

Anni [7]

Step-by-step explanation:

It is given that,

Initial speed of the car, u = 40 m/s

Final speed of the car, v = 25 m/s

Time taken, t = 3.5 s

(a) We need to find the magnitude and direction of the car’s acceleration as it slows down. It can be calculated using formula as :

$a=\dfrac{v-u}{t}$

$a=\dfrac{25-40}{3.5}$

$a=-4.28\ m/s^2$

The acceleration is in the opposite direction of motion i.e. west.

(b) Let s is the distance the car travel in the 3.5-s time period. It can be calculated using the third equation of motion as

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{25^2-40^2}{2\times (-4.28)}$

s = 113.9 meters

Hence, this is the required solution.

6 0

3 years ago

An oil tank has to be drained for maintenance. The tank is shaped like a cylinder that is

Dmitry_Shevchenko [17]

Fidn volume
V=hpir^2
d/2=r
1.4/2=0.7=r
height=3.5
V=3.5*3.14*0.7^2
V=5.3851

1.5 times t=5.3851
divide both sides by 1.5
t=3.59
round
t=4

4 minutes

6 0

3 years ago