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3 years ago

Juan construyó una rampa que tiene 5 m de largo y 1 m de altura. ?Cuánto mide la distancia (d) que recorre al subir la rampa

Mathematics

1 answer:

satela [25.4K]3 years ago

7 0

Answer:

$d=\sqrt{26}\ m$ or $d=5.1\ m$

Step-by-step explanation:

<em>The question in English is</em>

Juan built a ramp that is 5 m long and 1 m high. How much is the distance (d) he travels when he climbs the ramp?

let

x -----> the length of the ramp

y ----> the height of the ramp

d ----> the distance Juan travels when he climbs the ramp

we know that

Applying the Pythagoras Theorem

$d^{2}=x^{2}+y^{2}$

we have

$x=5\ m$

$y=1\ m$

substitute the given values

$d^{2}=5^{2}+1^{2}$

$d^{2}=26$

$d=\sqrt{26}\ m$ -----> exact value

$d=5.1\ m$ ----> approximate value

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Can someone help me with this thanks

Ugo [173]

The length of the brace required is 4.3m

What is sine rule?

In a ΔABC a, b and c are the sides and A, B and C are angles then,

$\frac{a}{SinA}=\frac{b}{sinB}=\frac{c}{sinC}$

We can find the length, l as shown below:

Let AB=3m, BC=2m and AC=l

Let ∠A=25°

So, in ΔABC

$\frac{BC}{sinA}=\frac{AB}{sinC}=\frac{AC}{SinB}$

$\frac{BC}{sinA}=\frac{AB}{sinC}$

$\frac{2}{sin25}=\frac{3}{sinC}$

$\angle{C}=sin^{-1}(\frac{3\times sin25}{2} )$

∠C=39.34°

∠A+∠B+∠C=180°

∠B=180°-25°-39.34°

∠B=115.66°

$\frac{BC}{sinA}=\frac{AC}{SinB}$

$\frac{2}{sin25}=\frac{l}{sin(115.66)}$

$l=\frac{2\times sin(115.66)}{sin25}$

l=4.2659

Rounding to nearest tenth of meter.

l=4.3m

Hence, the length of the brace required is 4.3m

Learn more about Sine Rule here:

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PLS HELP TIMED QUIZ The size of a television screen is given as the length of the diagonal. If a television is advertised as hav

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Answer:

D.24

Pythagorean states:

hypotenuse^2 = side1^2 + side2^2

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side2^2 = 900 - 324

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side2 = sqr root 576

side2 = 24

Source: http://www.1728.org/pythgorn.htm

Step-by-step explanation:

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Sergio039 [100]

Answer:

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Step-by-step explanation:

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4y - 2 (1 - y) = -44

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