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SashulF [63]
3 years ago
8

Question is attached, ill mark brainliest

Mathematics
2 answers:
Alborosie3 years ago
4 0

Answer:

w=7

Step-by-step explanation:

3w+4w=49

7w=49

w=49/7

w=7

umka2103 [35]3 years ago
4 0

Answer:

W is 7

Step-by-step explanation:

7x3 is 21 and 7x4 is 28 so 21+28=49

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What would 0.5598 × 908÷66 be
Andru [333]
It is 7.70 that is what I got
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3 years ago
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Explain, using an example, why need to multiply when converting from a larger unit to a smaller unit?
sveta [45]

Answer: Because you need more of a smaller unit to equal a larger unit.

Example:

If you want to convert 5km to meters, there must be more meters than kilometers.

5km(\frac{1000m}{1km} )=5000m

3 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
you have a 5-inch by 7-inch picture that you framed. When you hung the picture, you found that the entire area is 69.56 in. What
Montano1993 [528]

Answer:

d. 2.4 in

Step-by-step explanation:

So initially w/o frame the area was 5 x 7 = 35 sq in.

Now we have to add the width of the frame to both dimensions.

(5+w)(7+w) = 69.56 sq in

35 + 7w + 5w + w² = 69.56

w² + 12w + 35 = 69.56

w² + 12w - 34.56 = 0

I use the quadratic formula to solve this (x = -b±√b²-4ac / 2a). I cheat by using a quadratic program on the calculator :,)

w = 2.4

w = -14.4

Since we can't have a negative width, the answer must be <u>w = 2.4 inches.</u>

You can also just plug the answer choices one-by-one into the calculator with guess-and-check because this is multiple choice.

4 0
3 years ago
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Let m be the side length of the fourth smallest possible square with integer value of its area, and let n be the sum of all prim
Natali [406]

Answer:

510

Step-by-step explanation:

It's assumed 0 is not an integer.

M is obviously 2. The area of a square is the square of its side length. If this area must be an integer, then it must be the fourth integer, which is 4. The side length, M=\sqrt{4} = 2

A prime number is a positive integer which can be divided exactly by only itself and 1. In fact, it has only those two factors. The last statement excludes 1 as a prime number. Hence, the prime numbers less than 10 are 2, 3, 5 and 7. Their sum is N=2+3+5+7=17.

To evaluate MN(N-M),

2\times17(17-2)=34\times15=510

6 0
3 years ago
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