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satela [25.4K]
3 years ago
8

A circus performer walks on a tightrope 25 feet above the ground. The tightrope is supported by two beams and two support cables

. If the distance between each beam and the base of its support cable is 15 feet, what is the length of the support cable? Round to the nearest foot.
Mathematics
1 answer:
miv72 [106K]3 years ago
5 0
The length of the support cable is about 29 feet.

If you sketch out the question, you will see that it will form a right triangle with the cable as the hypotenuse.

Just set up and use the Pythagorean Theorem.

15^2 + 25^2 = c^2
850 = c^2
29.15 = c
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y=3x+2................(1)

y=x-4......................(11)

Subtract (11) from (1)

y-y=3x-x+2-(-4)

0=2x+2+4

0=2x+6

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3 years ago
A 10-foot ladder leans against a wall with its foot braced 3 feet from wallÍs base. How far up the wall does the ladder reach?
iren2701 [21]
The answer is √91 feet (which is the same as 9.54 feet).

Imagine this as a right triangle. A length of the foot is actually a hypotenuse (c). The distance from wall's base to the ladder foot is one of the sides of the triangle (let it be a).
So, using the Pythagorean theorem:
c² = a² + b²
It is given:
c = 10 feet
a = 3 feet
b = ?
     c² = a² + b²
⇒ 10² = 3² + b²
    100 = 9 + b²
    b² = 100 - 9 = 91
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PLEASE HELP ASAP -- WITH THE WORKK<br><br>50 points
MAXImum [283]

Part-A:-

<h3>(f×g)=</h3>

\\ \sf\longmapsto x^2+3x-70(4x-12)

\\ \sf\longmapsto 4x^2+12x-280-12x^2+36x+840

\\ \sf\longmapsto -8x^2+48x+560

Now

\\ \sf\longmapsto (f.g)(3)

\\ \sf\longmapsto -8(3)^2+48(3)+560

\\ \sf\longmapsto -72+144+560

\\ \sf\longmapsto 72+560

\\ \sf\longmapsto 632

Part-B:-

\\ \sf\longmapsto f(3)

\\ \sf\longmapsto (3)^2+3(3)-70

\\ \sf\longmapsto 9+9-70

\\ \sf\longmapsto 18-70

\\ \sf\longmapsto -52

Now

\\ \sf\longmapsto g(f(3))

\\ \sf\longmapsto g(-52)

\\ \sf\longmapsto 4(-52-3)

\\ \sf\longmapsto 4(-55)

\\ \sf\longmapsto -220

Part:-C

\\ \sf\longmapsto g(-2)

\\ \sf\longmapsto 4(-2-3)

\\ \sf\longmapsto 4(-5)

\\ \sf\longmapsto -20

Now

\\ \sf\longmapsto f(g(-2))

\\ \sf\longmapsto f(-20)

\\ \sf\longmapsto (-20)^2+3(-20)-70

\\ \sf\longmapsto 400-60-70

\\ \sf\longmapsto 400-130

\\ \sf\longmapsto 270

Part-D:-

  • g×f=f×g

\\ \sf\longmapsto (g.f)(-2)

\\ \sf\longmapsto -8(-2)^2+48(-2)+560

\\ \sf\longmapsto 32-96+560

\\ \sf\longmapsto -64+560

\\ \sf\longmapsto 496

6 0
2 years ago
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