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grandymaker [24]
3 years ago
13

The distance formula can be used to prove a quadrilateral has select one:

Physics
2 answers:
nadya68 [22]3 years ago
6 0
The answer is D
Hope this helped 
dybincka [34]3 years ago
4 0

Answer:  d. congruent sides

Explanation:

The distance formula is a formula in geometry to calculate the length of a line segment having two distinct point on a Cartesian plane

The distance formula to calculate the distance between two distinct points A(a,b) and B(p,q) is given by :-

\text{d}=\sqrt{(p-a)^2+(q-b)^2}

Since it is used to calculate the distance between two points therefore, it is used to check whether a quadrilateral has congruent sides.

Hence, the distance formula can be used to prove a quadrilateral has <u>congruent sides</u>.

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The positive end of battery 1 touching the negative end of battery 2 and wires connecting the negative of battery 1 to the light light bulb and the positive of battery 2 to the light bulb.

Explanation:

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A heavy boy and a lightweight girl are balanced on a mass-less seesaw. If they both move forward so that they are one-half their
Rom4ik [11]

Answer:

b) Nothing will happen,  the sea saw will still be balanced.

Explanation:

b) Nothing will happen,  the sea saw will still be balanced.

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3 years ago
Read 2 more answers
A 3.00 kg mud ball has a perfectly inelastic collision with a second mud ball that is initially at rest. the composite system mo
viva [34]
Perfectly inelastic collision is type of collision during which two objects collide, stay connected and momentum is conserved. Formula used for conservation of momentum is:

m_{1} * v_{1} +  m_{2} * v_{2} =  m_{1} * v'_{1} +  m_{2} * v'_{2}

In case of perfectly inelastic collision v'1 and v'2 are same.

We have following information:
m₁=3 kg
m₂=? kg
v₁=x m/s
v₂=0 m/s
v'1 = v'2 = 1/3 * v₁

Now we insert given information and solve for m₂:
3*v₁ + 0*? = 3*1/3*v₁ + m₂*1/3*v₁
3v₁ = v₁ + m₂*1/3*v₁
2v₁ = m₂*1/3*v₁
2 = m₂*1/3
m₂= 6kg

Mass of second mud ball is 6kg.
7 0
3 years ago
The graph represents the force applied on an 3.00kg crate while it moved 5.0m. A. How much total work is done on the crate? B. I
jeka57 [31]

a. We can calculate the amount of work by calculating the area under the graph.

first area (rectangular): 2.5 x 6 = 15

second area(trapezoid): 1/2 x (6+10) x 2.5 =20

total work done: 35 J

b. the force was first applied = 6 N

F = m.a

a = 6 : 3 = 2 m/s²

vf²=vi²+2as

vf²=6²+2.2.5

vf²=56

vf=7.5 m/s

8 0
3 years ago
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