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aev [14]
3 years ago
14

A wire is attached at the top of a 500 ft vertical antenna and is staked in the ground 1200 ft from the base of the antenna. how

long in feet is the guy wire? (enter an exact number.)
Mathematics
1 answer:
Pie3 years ago
4 0
A^2 + b^2 = c^2
(500)^2 + (1200)^2 = c^2
1,690,000 = c^2
take the square root of 1,690,000
Answer is 1,300 feet
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schepotkina [342]

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6.96%

Step-by-step explanation:

Assuming that the total number of races who did not finish the race are the ones who gave up and were disqualified:

(14+4)/230 = 0.06957

convert to percentage 0.06956*100= 6.957%

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Can someone help me Simplify: –(–x + 1)
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Since we have a negative on the outside of the parenthesis, we can just say that this is -1. So now we have -1(-x + 1). when multiply a negative by a negative it cancels out into a positive. and multiplying a negative with a positive will give you a negative. So now we have x - 1. 
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3 years ago
1. Suppose that the cost of producing x tablets is defined by
vredina [299]

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we are given

part-A:

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C(x) is cost of producing x tablets

so, vertical box is C(x)

so, we write in vertical box is "cost of producing x tablets"

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...........Answer

part-c:

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4 0
3 years ago
at 6:00 a.m., the temperature is 58°. At 2:00 p.m., temperature is 76°. Find the rate of change in degrees per hour during this
lesya [120]

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Step-by-step explanation:

The rate of change in degrees per hour is given by by the change in temperature divided by the change in time(by number of hours).

In this question:

From 6 am to 2 pm, there are 12 - 6 + 2 - 0 = 6 + 2 = 8 hours

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6 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.

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