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3 years ago

Please answer #1 and #2

Mathematics

1 answer:

DIA [1.3K]3 years ago

6 0

1)
x^2 + (3y/2z) = 7
2x^2z + 3y = 14z
3y = 14z - 2x^2z
3y = 2z(7 - x^2)
y = 2/3(z)(7 - x^2)

2)
(3zx^4) /(5+z) = 2y
3zx^4 = 2y(5+z)
3zx^4 = 10y + 2yz

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3 years ago

Assume you spend $275 for your course and you spend 10 weeks on your coursework. How much does it cost per hour to the nearest c

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<h2>Answer:</h2>

<em>The course costs, approximately, </em><u><em>$0.16 per hour</em></u><em>.</em>

<h2>Step-by-step explanation:</h2>

<h3>1. Convert the 10 weeks to hours.</h3>

10 weeks * 7= 70 days

70 days * 24= 1680 hours

<h3>2. Calculate cost/hour.</h3>

<em>Establish a rule of 3.</em>

<em>1680 hours -----> $275</em>

<em>x= (1 hour * $275) / 1680 hours</em>

<em>x= 0.1636904762 ---> Round to the nearest cent.</em>

<em>x= 0.16 ---> Final result.</em>

<em />

<h3>3. Conclude.</h3>

<em>The course costs, approximately, </em><u><em>$0.16 per hour</em></u><em>.</em>

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2 years ago

a boy is standing on a pole of height 14.7m throws a stone upwards. it moves in a vertical line slightly away from the pole and

fomenos

Answer:

The time taken for the upward motion is 1 second. The same time is taken for the downward motion

It reaches a maximum height of 4.9 meters.

Step-by-step explanation:

The equation of motion is:

$x(t) = -4.9t^{2} + 9.8t$

Since the term which multiplies t squared is negative, the graph is concave down, that is, x increases until the vertex, where it reaches it's maximum height, then it decreases.

Vertex of a quadratic equation:

Quadratic equation in the format $x(t) = at^{2} + bt + c$

The vertex is the point $(t_{v}, x(t_{v}))$ , in which

$t_{v} = -\frac{b}{2a}$

In this question:

$x(t) = -4.9t^{2} + 9.8t$

So $a = -4.9, b = 9.8$

Vertex:

$t_{v} = -\frac{9.8}{2*(-4.9)} = 1$

The time taken for the upward motion is 1 second.

$x(t_{v}) = x(1) = 9.8*1 - 4.9*(1)^{2} = 4.9$

It reaches a maximum height of 4.9 meters.

Downward motion:

From the vertex to the ground.

The ground is t when x = 0. So

$-4.9t^{2} + 9.8t = 0$

$4.9t^{2} - 9.8t = 0$

$4.9t(t - 2) = 0$

$4.9t = 0$

$t = 0$

$t - 2 = 0$

$t = 2$

It reaches the ground when t = 2 seconds.

The downward motion started at the vertex, when t = 1.

So the duration of the downward motion is 2 - 1 = 1 second.

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3 years ago