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timofeeve [1]
3 years ago
10

What additional information do you need to prove ∆ABC ≅ ∆DEF by the SAS Postulate?

Physics
1 answer:
miv72 [106K]3 years ago
6 0

Answer:

Option A

You need a Angle C congruent to angle F

Explanation:

EX) Side angle Side = sas

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A train travels 77 kilometers in 4 hours, and then 76 kilometers in 2 hours. What is its average speed?
Masja [62]

( (77/4) + 76/2 )/2 = 28.625 km/h is what i got

4 0
3 years ago
What is the acceleration of a dog that runs from 3 m/s to 6 m/s over a distance of 90m?
KonstantinChe [14]

Answer:

solution given:

acceleration (a)=?

initial velocity (u)=3m/s

final velocity (v)=6m/s

distance (s)=90m

we have

v²=u²+2as

substituting value

6²=3²+2*a*90

36=9+180a

36-9=180a

a=25/180

<u>a=0.1388m/s²</u>

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2 years ago
What heavy element was the chernobyl nuclear accelarators supposed to make?​
Gala2k [10]

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2 years ago
The speed of the ball when it returns to the same horizontal level
stellarik [79]

8.0 m/s if there is no air resistance. (B)

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4 0
3 years ago
A small logo is embedded in a thick block of crown glass (n = 1.52), 4.70 cm beneath the top surface of the glass. The block is
harkovskaia [24]

The concept required to solve this problem is the optical relationship that exists between the apparent depth and actual or actual depth. This is mathematically expressed under the equations.

d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})

Where,

d_g = Depth of glass

n_w = Refraction index of water

n_g = Refraction index of glass

n_{air} = Refraction index of air

d_w = Depth of water

I enclose a diagram for a better understanding of the problem, in this way we can determine that the apparent depth in the water of the logo would be subject to

d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})

d'w = (1.7cm) (\frac{1}{1.33})+(4.2cm)(\frac{1}{1.52})

d'w = 4.041cm

Therefore the distance below the upper surface of the water that appears to be the logo is 4.041cm

3 0
3 years ago
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