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Leviafan [203]
3 years ago
8

PZ HURRY. find angle b.​

Mathematics
2 answers:
sergij07 [2.7K]3 years ago
5 0

60° if im not mistaken. the right angle is 90° and 30°+b is also 90° as 180°-90°=90°. therefore 90°-30°=60°. hope it helped!

natita [175]3 years ago
4 0

Answer:

<u>b = 60°</u>

Step-by-step explanation:

If you look at the side under the 30° angle, there's a small red box, signifying a right (90°) angle. Doing the math, then, you would do:

90 = 30 + b

b = 60

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Solve 3x/5=1/6 <br> can someone help pls i dont get it
Kamila [148]

Answer:

x=5/18

Step-by-step explanation:

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2 years ago
3 - x/2 = 6 (Solve each equation) Check your answer
Yanka [14]
Answer to question
3-x/2=6
Subtract 3 from both sides
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Multiply 3 by -2
Final Answer: x=-6

Work check
3-x/2=6
Substitute -6 for x since you solved for x in the last equation
3-(-6)/2
Convert 3-(-6/2) to 3+6/2
3+6/2
Divide 6 by 2
3+3
Add
Final Answer: 6 (This 6 is the same 6 we used to solve for x in the first equation).
6 0
3 years ago
g Let G be a not necessarily abelian group with normal subgroups H and K such that H contains K (i.e., K ✂ G, H ✂ G, K ≤ H) and
allsm [11]

Answer:

Lets a,b be elements of G. since G/K is abelian, then there exists k ∈ K such that ab * k = ba (because the class of ab, [ab]_K is equal to [ba]_K, thus ab and ba are equal or you can obtain one from the other by multiplying by an element of K.

Since K is a subgroup of H, then k ∈ H. This means that you can obtain ba from ab by multiplying by an element of H, k. Thus, [ab]_H = [ba]_H . Since a and b were generic elements of H, then H/G is abelian.

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3 years ago
you work at a hair salon and get left a 35 tip on a 112 haircut and color service . What percent tip did your customer give you
levacccp [35]
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3 years ago
I need these from least to greatest A.11^4/11^11 B.1/11^-4 C.(11^-3)^-3 D.11^5•11^2
Sliva [168]

Answer:

\frac{ {11}^{4} }{ {11}^{11} } < \frac{1}{ {11}^{ - 4} }   < {11}^{5} . {11}^{2}  <  ( {11}^{ - 3} )^{ - 3}

Step-by-step explanation:

A) \frac{ {11}^{4} }{ {11}^{11} }  =  {11}^{4 - 11}  =  {11}^{ - 7}  \\  \\  B)\frac{1}{ {11}^{ - 4} }  =  {11}^{4}  \\  \\C) ( {11}^{ - 3} )^{ - 3} =  {11}^{ - 3( - 3)}  =  {11}^{9}   \\  \\  D){11}^{5} . {11}^{2}   =  {11}^{5 + 2}  =  {11}^{7}  \\  \\  \because \:  {11}^{ - 7}  <  {11}^{4}  <  {11}^{7}  <  {11}^{9}  \\  \\  \therefore \: \frac{ {11}^{4} }{ {11}^{11} } < \frac{1}{ {11}^{ - 4} }   < {11}^{5} . {11}^{2}  <  ( {11}^{ - 3} )^{ - 3}

6 0
3 years ago
Read 2 more answers
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