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kotegsom [21]
3 years ago
11

If a vehicle travels a distance of 5t^4-10t^2+6 miles in t+2 minutes what is the vehicles speed

Mathematics
1 answer:
satela [25.4K]3 years ago
3 0

speed = distance/time
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Subtract. <br> –84 – 55<br> a. –29<br> b. –139<br> c. 29<br> d. 139
Hitman42 [59]
-84 - 55 = -139 Here both have same sign (-) so u add and put the common sign(-) .... So the answer is option B ( -139)... Hope it helps!!!
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Which expressions are equivalent to the one below? Check all that apply. 53 • 5x
weeeeeb [17]
<span>

</span>Calculate 3 x 5, which is 15.Since 15 is two-digit, we carry the first digit 1 to the next column.

Calculate 5 x 5, which is 25. Now add the carry digit of 1, which is 26.Since 26 is two-digit, we carry the first digit 2 to the next column.

Bring down the carry digit of 2.


<span>Therefore, 53 x 5 = 265.</span>

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3 years ago
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Selena drove at an average speed of 50.55 Miles per hours for 1.75 hours. She stopped at a rest stop and then drove at an averag
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The correct answer would be 13.2375.
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Show that every triangle formed by the coordinate axes and a tangent line to y = 1/x ( for x &gt; 0)
vfiekz [6]

Answer:

Step-by-step explanation:

given a point (x_0,y_0) the equation of a line with slope m that passes through the  given point is

y-y_0 = m(x-x_0) or equivalently

y = mx+(y_0-mx_0).

Recall that a line of the form y=mx+b, the y intercept is b and the x intercept is \frac{-b}{m}.

So, in our case, the y intercept is (y_0-mx_0) and the x  intercept is \frac{mx_0-y_0}{m}.

In our case, we know that the line is tangent to the graph of 1/x. So consider a point over the graph (x_0,\frac{1}{x_0}). Which means that y_0=\frac{1}{x_0}

The slope of the tangent line is given by the derivative of the function evaluated at x_0. Using the properties of derivatives, we get

y' = \frac{-1}{x^2}. So evaluated at x_0 we get m = \frac{-1}{x_0^2}

Replacing the values in our previous findings we get that the y intercept is

(y_0-mx_0) = (\frac{1}{x_0}-(\frac{-1}{x_0^2}x_0)) = \frac{2}{x_0}

The x intercept is

\frac{mx_0-y_0}{m} = \frac{\frac{-1}{x_0^2}x_0-\frac{1}{x_0}}{\frac{-1}{x_0^2}} = 2x_0

The triangle in consideration has height \frac{2}{x_0} and base 2x_0. So the area is

\frac{1}{2}\frac{2}{x_0}\cdot 2x_0=2

So regardless of the point we take on the graph, the area of the triangle is always 2.

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