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kotegsom [21]
3 years ago
11

The escalator rises 25° from the ground and a vertical distance of 42 feet. How much horizontal distance does the escalator cove

r? Round to the nearest tenth.
Mathematics
1 answer:
zvonat [6]3 years ago
5 0
Tangent (25) = opp / adj = 42 / horizontal distance
horizontal distance =  42 / tangent (25)
horizontal distance = 42 / 0.46631
horizontal distance = <span> <span> <span> 90.068838326435200 </span> </span> </span>
horizontal distance = 90.1 feet


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Answer:

X=35

Step-by-step explanation:

Those 2 angles equal each other, meaning they're the same.

Knowing that you put them to equal each other.

2x+15=3x-20

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20+15=3x-2x

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Two circles with different radii have chords AB and CD, such that AB is congruent to CD. Are the arcs intersected by these chord
emmainna [20.7K]

The arcs intersected by these chords are not congruent.

Given that two circles with different radii have chords AB and CD, such that AB is congruent to CD.

Let r₁ and r₂ be the radii of two different circles with centers O and O' respectively.

Assuming that the each of the ∠АОВ  and ∠CO'D is less than or equal to π.

Then, we have isosceles triangle AOB and CO'D such that,

AO = OB = r₁,

CO' = O'D = r₂,

Let us assume that r₁< r₂;

We can see that arc(AB) intersected by AB is greater than arc(CD), intersected by the chord CD;

arc(AB) > arc(CD)      .......(1)

Indeed,

arc(AB) = r₁ angle (AOB)

arc(CD) = r₂ angle (CO'D)

So, we have to prove that ;

∠AOB >∠CO'D       ......(2)

Since each angle is less than or equal to π, and so

∠AOB/2  and ∠CO'D/2 is less than or equal to π

it suffices to show that :

tan(AOB/2) >tan(CO'D/2) ......(3)

From triangle AOB :

tan(AOB/2) = AB/(2*r₁)

tan(CO'D/2) = CD/(2*r₂)

Since AB = CD and r₁ < r₂ (As obtained from the result of (3) ), therefore, arc(AB) > arc(CD).

Hence, for two circles with different radii have chords AB and CD, such that AB is congruent to CD but the arcs intersected by these chords are not congruent.

Learn more about congruent from here brainly.com/question/1675117

#SPJ1

6 0
2 years ago
Solveing systems of equations by substitution y=6x-11 -2x-3y=-7
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\bf \begin{cases} \boxed{y}=6x-11\\ \cline{1-1} -2x-3y=-7 \end{cases}\qquad \implies \stackrel{\textit{substituting \underline{y} in the 2nd equation}}{-2x-3\left( \boxed{6x-11} \right)=-7}

\bf -2x-18x+33=-7\implies -20x+33=-7\implies -20x=-40 \\\\\\ x=\cfrac{-40}{-20}\implies \blacktriangleright x=2 \blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{since we know that }}{y=6x-11}\implies y=6(2)-11\implies y=12-11\implies \blacktriangleright y=1 \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (2,1)~\hfill

7 0
3 years ago
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