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Lapatulllka [165]
2 years ago
9

A linear function has a slope of 8/3 and passes through the point (0,5). What is the equation of the line?

Mathematics
2 answers:
MArishka [77]2 years ago
3 0

This problem already gave us the slope and y-intercept, so all that's left to do is plug those two values into slope-intercept form.

Slope-Intercept Form: y = mx + b

---m is the slope

---b is the y-intercept

y = 8/3x + 5

Hope this helps!! :)

jok3333 [9.3K]2 years ago
3 0

y = 8/3x + 5

hope this helps !! :)

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Elis [28]
1.5b=11-9
1.5b=2
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3 years ago
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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

Or

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
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Answer:

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Step-by-step explanation:

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\bf slope = m = \cfrac{rise}{run} \implies \cfrac{ f(x_2) - f(x_1)}{ x_2 - x_1}\impliedby \begin{array}{llll} average~rate\\ of~change \end{array}\\\\[-0.35em] \rule{34em}{0.25pt}\\\\ f(x)= 2x^2-1\qquad \begin{cases} x_1=0\\ x_2=5 \end{cases}\implies \cfrac{f(5)-f(0)}{5-0} \\\\\\ \cfrac{[2(5)^2-1]~~-~~[2(0)^2-1]}{5}\implies \cfrac{50-(-1)}{5}\implies \cfrac{50+1}{5}\implies \cfrac{51}{5}\implies 10\frac{1}{5}

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