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inessss [21]
3 years ago
14

Find the volume of the composite figure formed by a hemisphere with the radius

Mathematics
1 answer:
LekaFEV [45]3 years ago
5 0

Answer:

8209.44

Step-by-step explanation:

CUBE: 20x20x20=8000

HEMISPHERE=2/3\pir²=

2/3\pi(10)²=209.44

8000+209.44=8209.44

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HELP PLEASE <br><br>must show work ​
matrenka [14]

Answer:

1. 4n^3

2. 4k^7

3. 3

4. -30x

5. -6

Step-by-step explanation:

1. The prime factorization of 12 is 2 x 2 x 3 and the prime factorization of 16 is 2 x 2 x 2 x 2. When you look at these two expressions you can see the common factors of these two numbers are 2 x 2, which is 4. Next, we look at the GCF of the N's which would be n^3 since n^5 has three N's in it. Therefore, we get 4n^3 when we multiply the two together.

2. The factors of 8 are 1, 2, 4, and 8. Out of these, 1, 2, and 4 are the only factors that 20 shares with it and 4 is the greatest. Then, we look at the K's and the GCF of the K's is k^7 since k^8 has seven K's. We multiply the two and we get 4k^7.

3. Since one of the numbers of the three given here does not include the variable n, there will not be any N's in the GCF of the three, so we don't have to worry about that. Now, we just find the GCF of 18, -24, and -21. The factors of 18 are 1, 2, 3, 6, 9, and 18, the factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24, and lastly, the factors of 21 are 1, 3, 7, and 21. From these, 3 is the biggest common divisor, therefore the GCF is 3.

4. Between the two X's, X^1 is the biggest amount of X's this GCF has, so the final GCF will be some constant multiplies with X. Since we are dealing with bigger numbers on this problem, we should use prime factorization. The prime factorization of 90 is 2 x 3 x 3 x 5, and the prime factorization of 120 is 2 x 2 x 2 x 3 x 5. From these expressions, we take the biggest amount of each common factor as we can. Since these expressions both have 2, we take the smaller amount of 2's which is one two. Then we get one three from both expressions, and one five as well. 2 times 3 times 5 equals 30, therefore, we get -30x, and not 30x, because both of these numbers are negatives.

5. All of these numbers do not have an x, so there won't be an x in our GCF. Another method of quickly finding the GCF of numbers is to look at the smallest number's factors first to see what factors it shares with the other numbers. The factors of 12 are 1, 2, 3, 4, 6, and 12. 42 and 30 do not have the factor 12, so we can go down the list and see if 42 and 30 share the factor 6, which they do since 6 times 7 is 42 and 6 times 5 is 30. Since all of these numbers share the negative sign, the GCF of these three numbers is -6.  

5 0
3 years ago
(12x5+3x25)-8<br><br> 12x5=60<br> 3x25=75<br> 60+75=135<br> 135-8=127
lions [1.4K]

Answer:

127

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
When you divide , we are
fomenos
Could you be a little more specific.
5 0
3 years ago
Read 2 more answers
Solve the equation. And write all solutions in general form.
Amiraneli [1.4K]

Answer:

x = pi/2 + 2 pi n                            x = pi + 2 pi n   where n is an integer

x = 5pi /3 + 2 pi n                      

Step-by-step explanation:

8 cos^2 x + 4 cos x-4 = 0

Divide by 4

2 cos^2 x +  cos x-1 = 0

Let u = cos x

2 u^2 +u -1 =0

Factor

(2u -1) ( u+1) = 0

Using the zero product property

2u-1 =0    u+1 =0

u = 1/2      u = -1

Substitute cosx for u

cos x = 1/2    cos x = -1

Take the inverse cos on each side

cos ^-1(cos x) = cos ^-1(1/2)   cos ^-1( cos x) =cos ^-1( -1)

x = pi/2 + 2 pi n                            x = pi + 2 pi n   where n is an integer

x = 5pi /3 + 2 pi n                      

8 0
3 years ago
F and H are sets of real numbers defined as follows.
Marina CMI [18]

Answer:

(a)F \cup H=(3, \infty)\\(b)F \cap H=(6, \infty)

Step-by-step explanation:

Given the sets F and H of real numbers defined as given:

F={w | w>3}

H={w | w>6}

To write a set in interval  notation, we use these brackets, (  ) for < or > while we use [ or ] for \leq $ or \geq.

Since our sets are defined by strict inequalities(< or >)

In interval notation

F=(3,\infty), H=(6,\infty)

(a) F \cup H =(3,\infty) \cup (6,\infty) \\\\F \cup H=(3, \infty)\\\\(b) F \cap H =(3,\infty) \cap (6,\infty) \\\\F \cap H=(6, \infty)

8 0
3 years ago
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