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3 years ago

9

the outdoor temperature was 8 degrees at midnight. The temperature declined 5 degree during each of the next 3 hours What was th

e temperature at 3am???? I NEED HELP

1 answer:

kompoz [17]3 years ago

4 0

I would say it was 3 degrees at 3am

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I need help Someone please help me!!

Zanzabum

2. The area for a circle is as follows:
A = π ${r^2}$

The radius of the circle is 4 in, so we plug that into the equation.

A = π (4 ${in.^2}$ )

4 ${in.^2}$ = (4 in.)(4 in.) = 16 ${in.^2}$

A = 16π ${in.^2}$

This is the area for the whole pizza, so divide that by 2 to get the area of the pepperoni

a) 8π ${in.^2}$

For the bacon, we can just divide the value we already have by 2, since it is half the area of the pepperoni

b) 4π ${in.^2}$

3. This question is asking for the circumference, whose equation is as follows:
C = 2 π r

We can use the area formula, A = π ${r^2}$ to find the radius of this circle
36π = π ${r^2}$

Divide both sides by π:
36 = ${r^2}$

Square root both sides:
r = 6 in.

Plug in this value to the circumference formula, C = 2 π r
C = (2)(π)(6 in.)

Your final answer is:
C = 12π in.

3 0

4 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

Polynomial Long Division

Crazy boy [7]

Here is my answer. Hope it makes sense!

3 0

3 years ago

Kelly has made 16 out of 25 free throws. the probability that she will make her next free throw is 64%. Is this an Experimental

Anna71 [15]

Well percents are probability, and probability is similar to theory. So i think its theoritical probability.

Hope this helps.

7 0

4 years ago

5+656+-425*3/4<35-73*3/7

Komok [63]

The answer is false

if you solve both sides it says that 627.25 is less than 3.714 which isn't correct because 627.25 is obviously greater than 3.714

5 0

4 years ago