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3 years ago

6

The time a traffic light remains yellow is one second more than 0.05 times the speed limit. What is the yellow time for a traffi

c light on a street witha speed limit of 30 mph?

1 answer:

weqwewe [10]3 years ago

4 0

30 * 0.05 + 1 = X
1.5 + 1 = X
2.5 = X

The light remains yellow for 2.5 seconds on a street with a speed limit of 30.

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Find two numbers if their difference is 16 and their ratio is 5:7.

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Let us say that the numbers are x and y. The given clues are:

1. x – y = 16 (so x is the larger number)

2. y / x = 5 / 7

rewriting eqtn 2 in terms of y:

y = (5/7) x

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olga_2 [115]

The solution for the expression (v²)(m⁴)/(v²)(m³) will be m. The correct option is C.

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The given expression will be solved as below:-

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2 years ago

Find the equation, (f(x) = a(x - h)2 + k), for a parabola containing point (2, -1) and having (4, -3) as a vertex. What is the s

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Step-by-step explanation:

A parabola is written in the form

$f(x)=a((x-h)^2+k)$ (1)

where:

$h$ is the x-coordinate of the vertex of the parabola

$ak$ is the y-coordinate of the vertex of the parabola

$a$ is a scale factor

For the parabola in the problem, we know that the vertex has coordinates (4,-3), so we have:

$h=4$ (2)

$ak=-3$

From this last equation, we get that $a=\frac{-3}{k}$ (3)

Substituting (2) and (3) into (1) we get the new expression:

$f(x)=-\frac{3}{k}((x-4)^2+k) = -\frac{3}{k}(x-4)^2 -3$ (4)

We also know that the parabola contains the point (2,-1), so we can substitute

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So the equation of the parabola is:

$f(x)=\frac{1}{2}((x-4)^2 -6)$ (5)

Now we want to rewrite it in the standard form, i.e. in the form

$f(x)=ax^2+bx+c$

To do that, we simply rewrite (5) expliciting the various terms, we find:

$f(x)=\frac{1}{2}((x^2-8x+16)-6)=\frac{1}{2}(x^2-8x+10)=\frac{1}{2}x^2-4x+5$

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Answer:

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