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drek231 [11]
3 years ago
15

Please help. This is summer homework that's due in 2 days!

Mathematics
1 answer:
iVinArrow [24]3 years ago
4 0
Split the surface area into geometric shapes as shown in the figure below.

Top part
The area of one triangle is
A₁ = (1/2)*(10 cm)*(4 cm) = 20 cm²
There are 3 of these triangles.

Vertical sides..
The area of one vertical side is
A₂ = (10 cm)*(6 cm) = 60 cm²
There 3 of them.

Bottom part
The triangle is equilateral with each side  = 10 cm, and each angle = 60 deg.
The area is (from the Law of Sines)  equal to
A₃ = (1/2)(10 cm)*(10 cm)*sin(60) = 43.3 cm²

The total surface area is
3A₁ + 3A₂ + A₃
= 3*20 + 3*60 + 4.3
= 283.3 cm²

Answer: The surface area is  283.3 cm²

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Which function is defined for x=0
Sav [38]

0 would be on the top, hope this helps

4 0
3 years ago
What property was used to rewrite the polynomial expression (x+5)+3(a+(6x)y) to 5+(x+3a+(18x)y)
Umnica [9.8K]
<span>3(a+(6x)y) was clearly multiplied out as seen by the 3a and 18xy, so the distributive property was used there. In addition, the commutative and associative properties state that you can rearrange sums, so those were used too </span><span />
4 0
3 years ago
paris run 3 1/2 laps around the circular track in the school gym. The track has a diameter of 1/2 mile. About how far does paris
fiasKO [112]

Answer:

5.497787144 miles

Step-by-step explanation:

Remember that circumference of a circle can be calculated with:

2\pi r

r=radius

We also know that the radius is just a half of the diameter.

So, the radius of the track is (1/2)/2=1/4 miles

Now, use the circumference of a circle formula to find how much she runs in 1 lap.

2\pi (\frac{1}{4} )

Simplify

\frac{\pi }{2}

This is how much she runs in 1 lap.

Multiply by 3.5 for 3 and a half laps.

\frac{\pi }{2}*3.5

The answer is around 5.497787144 miles.


7 0
3 years ago
Given that , the _____ justifies the conclusion that ∆ABD ≅ ∆AEC.
trasher [3.6K]
AAS Postulate

It is given that CE = BD so we know "S" (representing side) has to be in the three letter postulate.
It is also given that angle DBA and angle CEA are right angles, so therefore they are congruent. Now we know that an "A" must also be in the postulate.

Lastly, we know that the triangles have a second angle, EAB, in common because they share it overlappingly. So there must be another "A" in the postulate. 

Now we need to look at the order in which it is presented. The order follows Angle, Angle, Side so the postulate must be the AAS postulate. Hope this helps!

3 0
3 years ago
Read 2 more answers
NO LINKS PLEASE I ACTUALLY NEED HELP THEY'RE 2 PARTS
eimsori [14]
I’m on the same question
3 0
3 years ago
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