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yawa3891 [41]
3 years ago
9

Correct answers will get brainliest

Mathematics
2 answers:
matrenka [14]3 years ago
7 0

Step-by-step explanation:

1 i dont know

2.

5(3y - 2) = 35 \\  = 15y - 10 = 35 \\  = 15y = 35 + 10 = 45 \\y  =  \frac{45}{15}  = 3

3.

\frac{88}{100}  =  \frac{22}{25}

kondor19780726 [428]3 years ago
4 0

Answer:

Step-by-step explanation:

I hope it could help u and if isn't clear then let me know ..

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Can some one do all these questions for me.
Zielflug [23.3K]
Which questions can you please send a link? :)
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3 years ago
Triangle ABC has these side measurements: AB = 17 BC = 18 AC = 21 Order the angles of the triangle from largest measure to small
vodka [1.7K]

The angles are

  • AC
  • BC
  • AB

Hope this helps :)

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3 years ago
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Oduvanchick [21]
The answer is 64 because they are alternate interior angles. Alternate interior angles are always congruent.
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3 years ago
What is the length of a diagonal of a cube with a side length of 8 inches?
sashaice [31]

Answer:

13.9 inches

Step-by-step explanation:

Let the length of diagonal of cube be d inches.

\therefore \: d = length \: of \: side \sqrt{3}  \\ \hspace{16 pt}  = 8 \sqrt{3}  \\ \hspace{16 pt} =8 \times  1.73205081 \\ \hspace{16 pt} = 13.8564065 \\    \huge \red{ \boxed{\therefore \: d = 13.9 \: inches }}\\

8 0
3 years ago
Read 2 more answers
Let H and K be subgroups of a group G, and let g be an element of G. The set <img src="https://tex.z-dn.net/?f=%5Cmath%20HgK%20%
34kurt

Answer:

Yes, double cosets partition G.

Step-by-step explanation:

We are going to define a <em>relation</em> over the elements of G.

Let x,y\in G. We say that x\sim y if, and only if, y\in HxK, or, equivalently, if y=hxk, for some h\in H, k\in K.

This defines an <em>equivalence relation over </em><em>G</em>, that is, this relation is <em>reflexive, symmetric and transitive:</em>

  • Reflexivity: (x\sim x for all x\in G.) Note that we can write x=exe, where e is the <em>identity element</em>, so e\in H,K and then x\in HxK. Therefore, x\sim x.
  • Symmetry: (If x\sim y then y\sim x.) If x\sim y then y=hxk for some h\in H and k\in K. Multiplying by the inverses of h and k we get that x=h^{-1}yk^{-1} and is known that h^{-1}\in H and k^{-1}\in K. This means that x\in HyK or, equivalently, y\sim x.
  • Transitivity: (If x\sim y and y\sim z, then x\sim z.) If x\sim y and y\sim z, then there exists h_1,h_2\in H and k_1,k_2\in K such that y=h_1xk_1 and z=h_2yk_2. Then, \\ z=h_2yk_2=h_2(h_1xk_1)k_2=(h_2h_1)x(k_1k_2)=h_3xk_3 where h_3=h_2h_1\in H and k_3=k_1k_2\in K. Consequently, z\sim x.

Now that we prove that the relation "\sim" is an equivalence over G, we use the fact that the <em>different equivalence classes partition </em><em>G.</em><em> </em>Since the equivalence classes are defined by [x]=\{y\in G\colon x\sim y\} =\{y\in G \colon y=hgk\ \text{for some } h\in H, k\in K \}=HxK, then we're done.

5 0
3 years ago
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