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olchik [2.2K]
3 years ago
8

Name 2 alternate exterior angles

Mathematics
1 answer:
Marizza181 [45]3 years ago
4 0

Answer:

When two lines are crossed by another line (called the Transversal): Alternate Exterior Angles are a pair of angles on the outer side of each of those two lines but on opposite sides of the transversal. In this example, these are two pairs of Alternate Exterior Angles: a and h.

Step-by-step explanation:

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I need help on 1-6 and If you can, can you explain how to do these? I'm not entirely sure how to.
gayaneshka [121]
1. Using c=2pi(r), plug in 7 for r and solve. Then using a=pi(r)^2, plug in 7 for r once again and solve.

2. First, the diameter (d) is 12 so to get the radius (r), divide 12 by 2 and you should get 6. Then use c=2pi(r) for circumference and a=pi(r)^2 for area to solve.

3. To get the area of the semicircle, divide 16 by 2 to get the radius (r), plug it into a=pi(r)^2, and divide the answer you get for a by 2. To get the area of the triangle, use a=1/2bh, plugging in 16 for b and 10 for h. Finally, add your two answers (the a's from the semicircle and triangle problems).

4. Multiply 20 by 5.5 to get the area of the triangle. Then multiply 4.5 by 20 to get the area of the parallelogram and add your two quotients.

5. Use a=1/2bh and plug in 4 for b and 3 for h and solve. Then multiply the quotient by 10 and there's your volume. To find the surface area, solve SA=(10×4)+(10×3)+(10×5)+12. All I did there was find the area of all the sides and added them together.

6. To find the triangle's volume, use a=1/2bh (b=4, h=1.5) and then multiply the quotient of that by 2.5. To find the rectangle's volume, use v=lwh (l=4, w=2.5, h=2) and solve. Finally, add the triangle's volume and the rectangle's volume to get the total volume. To get its surface area, start with the rectangle. Find the areas of all the sides and add them together but then subtract the 2.5×4 rectangle as it is not on the surface. It should look like this: SA=2(4×2)+2(2.5×2)+10. Again, all I did was find the areas of all the rectangle's sides on the surface and added them. Next, find the triangle's areas on the surface and it should look like this: SA=2(1.5×4)+2(2.5×2.5). Finally, add both values of SA from the triangle and rectangle and there's your surface area.
7 0
3 years ago
The area of ​​the bottom of the two cylindrical vessels A and B is 400cm ^ 2 and 300cm ^ 2, respectively. A jar is about 6cm hig
Sonbull [250]

Answer:

Water in container B will be at 6.own

Step-by-step explanation:

calculate the volume in cont. A

VoA=400×6

=2400cm cube

VoB=420ml

=420cm cube [1ml=1cm cube]

now add both volumes

V=2400+420

=2820cm cube

Now,

2820 cm cube in jar B so

2820=420×h

h=6.71

4 0
3 years ago
Simplify the expression (x^19 y^21)^4/(x^2 y^6)^2<br><br> The simplified expression is ________.
Masja [62]
I got x^72y^72 i hope this helps

3 0
2 years ago
Read 2 more answers
Dont get it wrong, if its wrong i fail, explain how you got your answer
damaskus [11]

Answer: 50.24

Area of a circle is pi times r^2

The r is 1/2 the d

So 8/2 = 4

Pi times 4^2= 50.24

7 0
3 years ago
If 28% of students in College are near-sighted, the probability that in a randomly chosen group of 20 College students, exactly
cluponka [151]

Answer: (D) 16%

Step-by-step explanation:

Binomial probability formula :-

P(x)=^nC_xp^x(1-p)^n-x, where n is the sample size , p is population proportion and P(x) is the probability of getting success in x trial.

Given : The proportion of students in College are near-sighted : p= 0.28

Sample size : n= 20

Then, the the probability that in a randomly chosen group of 20 College students, exactly 4 are near-sighted is given by :_

P(x=4)=^{20}C_4(0.28)^4(1-0.28)^{20-4}\\\\=\dfrac{20!}{4!16!}(0.28)^4(0.72)^{16}\\\\=0.155326604912\approx0.16\%

Hence, the probability that in a randomly chosen group of 20 College students, exactly 4 are near-sighted is closest to 16%.

7 0
3 years ago
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