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2 years ago

Ella completed the following work to test the equivalence of two expressions. '

Mathematics

1 answer:

Romashka [77]2 years ago

8 0

Answer:

The expressions are not equivalent because Ella did not know that you can’t use substitution to test for equivalence.

Step-by-step explanation:

Equivalent algebraic expressions are those expressions which on simplification give the same resulting expression.

Two algebraic expressions are said to be equivalent if their values obtained by substituting any values of the variables are same.

Two expressions 3f+2.6 and 2f+2.6 are not equivalent, because when f=1,

3f+2.6=3.1+2.6=3+2.6=5.6

2f +2.6=2.1+2.6=2+2.6=4.6

5.6≠ 4.6

Method of substitution can only help her to decide the expresssions are not equivalent, but if she wants to prove the expressions are equivalent, she must prove it for all values of f.

3f+2.6=2f+2.6

3f=2f

3f-2f=0

f=0

This is true only when f=0.

Hence,

The expressions are not equivalent because Ella did not know that you can’t use substitution to test for equivalence.

this what i know

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Answer:

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Step-by-step explanation

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2 years ago

El caño de una fuente está inclinado 60° sobre la horizontal. Si el agua sale del caño con una velocidad inicial de 10 m/s: a) ¿

jeka94

Answer:

A) Parabola; B) 3.83 m; C) 8.84 m; D) 10 m/s

Step-by-step explanation:

A) Shape of water jet

The water jet has the shape of a parabola.

B) Maximum height

Data:

θ = 60°

u = 10 m/s

a = 9.8 m·s⁻²

Calculations:

1. Calculate the horizontal and vertical components of the velocity

$u_{\text{h}} = u \cos \theta = \text{10 m/s} \times \cos 60 ^{\circ} = \text{10 m/s} \times 0.5 = \text{5 m/s}\\u_{\text{v}} = u \sin \theta = \text{10 m/s} \times \sin 60 ^{\circ} = \text{10 m/s} \times 0.866 = \text{8.66 m/s}$

2. Maximum height

$H = \dfrac{(u_{\text{v}})^{2}}{2a} = \dfrac{8.66^{2}}{2\times 9.8} =\textbf{3.83 m}$

C) Range

1. Calculate the time of flight

Use the vertical component of velocity to calculate the time to the maximum height of the stream.

$u_{\text{v}} = at\\t = \dfrac{ u_{\text{v}}}{a} = \dfrac{\text{8.66 m$\cdot$s}^{-1}}{\text{9.8 m$\cdot$s}^{-2}}= \text{0.884 s}$

It will take the same time to reach the ground.

Thus,

Time of flight = 2t = 2 × 0.884 s = 1.77 s

2. Calculate the horizontal distance

s = vt = 5 m·s⁻¹ × 1.77 s = 8.84 m

You should place the drain 8.84 m from the pipe.}

D) Modulus of velocity

The stream of water will hit the drain with the same velocity as when it left the pipe.

Thus, the modulus of the velocity is 10 m/s.

The graph below shows the trajectory of the water stream.

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