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hoa [83]
3 years ago
7

Simplify (rs) (25) -2 (47)4

Mathematics
1 answer:
Katen [24]3 years ago
5 0

Answer:

there should be multiplication sign...so value will be (-351)

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What answer this is ?
Umnica [9.8K]

Answer:

A and C

Step-by-step explanation:

If the fraction is negative either the numerator or the denominator can be marked negative and it stays the same

7 0
3 years ago
Find the inequality represented by the graph ​
goldfiish [28.3K]

Answer:

  • y > 1/3x - 3

Step-by-step explanation:

<u>Finding the line using the points:</u>

  • (0, -3) and (3, -2)

<u>The equation:</u>

  • y = mx + b

<u>The slope is:</u>

  • m = (-2 + 3)/(3-0) = 1/3
  • y- intercept is -3

<u>So the line is:</u>

  • y = 1/3x - 3

<u>As per the graph the line is dotted and the shaded area is above the line,. so the inequality is:</u>

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8 0
3 years ago
What is the formula of the curved surface of a cylinder?
VladimirAG [237]

Answer:

2πr × h = 2πrh

Step-by-step explanation:

that is the formula for a curved surface of a cylinder 2πr × h = 2πrh.

3 0
1 year ago
Marcie solved the following inequality, and her work is shown below.
mihalych1998 [28]

Answer:

She didn't change the direction of the sign

Step-by-step explanation:

When dividing both sides by a negative number in an inequality, you must change the sign.

4 0
3 years ago
Let H be a subgroup of a group G. We call H characteristic in G if for any automorphism σ∈Aut(G) of G, we have σ(H)=H.
choli [55]

Answer:Problem 1. Let G be a group and let H, K be two subgroups of G. Dene the set HK = {hk : h ∈ H,k ∈ K}.

a) Prove that if both H and K are normal then H ∩ K is also a normal subgroup of G.

b) Prove that if H is normal then H ∩ K is a normal subgroup of K.

c) Prove that if H is normal then HK = KH and HK is a subgroup of G.

d) Prove that if both H and K are normal then HK is a normal subgroup of G.

e) What is HK when G = D16, H = {I,S}, K = {I,T2,T4,T6}? Can you give geometric description of HK?

Solution: a) We know that H ∩ K is a subgroup (Problem 3a) of homework 33). In order to prove that it is a normal subgroup let g ∈ G and h ∈ H ∩ K. Thus h ∈ H and h ∈ K. Since both H and K are normal, we have ghg−1 ∈ H and ghg−1 ∈ K. Consequently, ghg−1 ∈ H ∩ K, which proves that H ∩ K is a normal subgroup.

b) Suppose that H G. Let K ∈ k and h ∈ H ∩ K. Then khk−1 ∈ H (since H is normal in G) and khk−1 ∈ K (since both h and k are in K), so khk−1 ∈ H ∩ K. This proves that H ∩ K K.

c) Let x ∈ HK. Then x = hk for some h ∈ H and k ∈ K. Note that x = hk = k(k−1hk). Since k ∈ K and k−1hk ∈ H (here we use the assumption that H G), we see that x ∈ KH. This shows that HK ⊆ KH. To see the opposite inclusion, consider y ∈ KH, so y = kh for some h ∈ H and k ∈ K. Thus y = (khk−1)k ∈ HK, which proves that KH ⊆ HK and therefoere HK = KH. To prove that HK is a subgroup note that e = e · e ∈ HK. If a,b ∈ HK then a = hk and b = h1k1 for some h,h1 ∈ H and k,k1 ∈ K. Thus ab = hkh1k1. Since HK = KH and kh1 ∈ KH, we have kh1 = h2k2 for some k2 ∈ K, h2 ∈ H. Consequently,

ab = h(kh1)k1 = h(h2k2)k1 = (hh2)(k2k1) ∈ HK

(since hh2 ∈ H and k2k1 ∈ K). Thus HK is closed under multiplication. Finally,

Step-by-step explanation:

6 0
3 years ago
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