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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.)

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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Out of 230 racers who started the marathon,

schepotkina [342]

Answer:

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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3 years ago

Can someone help me Simplify: –(–x + 1)

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3 years ago

1. Suppose that the cost of producing x tablets is defined by

vredina [299]

Answer:

we are given

part-A:

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

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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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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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2 years ago