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Flauer [41]
2 years ago
7

Please help with this math question

Mathematics
1 answer:
max2010maxim [7]2 years ago
8 0
Mmmm Hope this help you

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Tina is packing for a five-day trip. She has one blue, one red, one white, one green, and one grey t-shirt. Tina will not wear e
agasfer [191]

Answer:

This means that Tina has enough t-shirts for the entire trip and doesn't need to pack any more t-shirts.

Step-by-step explanation:

In total, she has five t-shirts, and she is packing for a five-day trip. Therefore, she has enough shirts to last her the entire trip.

5 0
2 years ago
Compute the probability of random sly selecting a queen or diamond.
FrozenT [24]
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3 0
3 years ago
The cost C (in dollars) to participate in a ski club is given by the literal equation C=85x+60, where X=C/85-12/17 is the number
dybincka [34]

Given:

The cost C (in dollars) to participate in a ski club is given by,

\begin{gathered} C=85x+60 \\ (or) \\ x=\frac{C}{85}-\frac{12}{17} \end{gathered}

Where x is the number of ski trips we take.

To find:

The number of ski trips when a total cost of $315 and $485 is spent.

Explanation:

Substituting C = 315 in the given equation we get,

\begin{gathered} x=\frac{315}{85}-\frac{12}{17} \\ x=\frac{315-60}{85} \\ x=\frac{255}{85} \\ x=3\text{ ski trips} \end{gathered}

Substituting C = 485 in the given equation we get,

\begin{gathered} x=\frac{485}{85}-\frac{12}{17} \\ x=\frac{485-60}{85} \\ x=\frac{425}{85} \\ x=5\text{ ski trips} \end{gathered}

Thus,

If we spend a total cost of $315, we can take 3 ski trips.

If we spend a total cost of $485, we can take 5 ski trips.

Final answer:

• If we spend a total cost of $315, we can take 3 ski trips.

,

• If we spend a total cost of $485, we can take 5 ski trips.

8 0
1 year ago
Write the following ratio in their simplest form (a)150minutes:1 1/2 hours (b) 150cm:1060mm ​
spayn [35]

Answer:

a) 5:3

b) 7.5:5.3

Step-by-step explanation:

150min : 90min (1 and a half hour)

10 : 6

5 : 3

150 × 10 mm : 1060 mm

1500 : 1060

750 : 530

7.5 : 5.3

7 0
2 years ago
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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