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2 years ago

7

5x-2y 2.4what is the constant

1 answer:

Veronika [31]2 years ago

4 0

For equation 5x - 2y = 2.4 the constant can be solve by converting the equation to a point slope form equation of a line. It is in the form y = mx + b

Dividing the whole equation by -2

<span>Y = 2.5x – 1.2 so the constant is 1.2</span>

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Find the exact length of the curve. x=et+e−t, y=5−2t, 0≤t≤2 For a curve given by parametric equations x=f(t) and y=g(t), arc len

Rama09 [41]

The length of a curve <em>C</em> parameterized by a vector function <em>r</em><em>(t)</em> = <em>x(t)</em> i + <em>y(t)</em> j over an interval <em>a</em> ≤ <em>t</em> ≤ <em>b</em> is

$\displaystyle\int_C\mathrm ds = \int_a^b \sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2} \,\mathrm dt$

In this case, we have

<em>x(t)</em> = exp(<em>t</em> ) + exp(-<em>t</em> ) ==> d<em>x</em>/d<em>t</em> = exp(<em>t</em> ) - exp(-<em>t</em> )

<em>y(t)</em> = 5 - 2<em>t</em> ==> d<em>y</em>/d<em>t</em> = -2

and [<em>a</em>, <em>b</em>] = [0, 2]. The length of the curve is then

$\displaystyle\int_0^2 \sqrt{\left(e^t-e^{-t}\right)^2+(-2)^2} \,\mathrm dt = \int_0^2 \sqrt{e^{2t}-2+e^{-2t}+4}\,\mathrm dt$

$=\displaystyle\int_0^2 \sqrt{e^{2t}+2+e^{-2t}} \,\mathrm dt$

$=\displaystyle\int_0^2\sqrt{\left(e^t+e^{-t}\right)^2} \,\mathrm dt$

$=\displaystyle\int_0^2\left(e^t+e^{-t}\right)\,\mathrm dt$

$=\left(e^t-e^{-t}\right)\bigg|_0^2 = \left(e^2-e^{-2}\right)-\left(e^0-e^{-0}\right) = \boxed{e^2-\frac1{e^2}}$

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den301095 [7]

Answer:

Step-by-step explanation:

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2 years ago

24. What are the x and y intercepts of the following

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Answer:

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B. x-intercept (-8,0) and y-intercept (0,6)wrong

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2 years ago

2/5x = 6<br> (One Step Equation)

BaLLatris [955]

Answer:

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x 5 x 5

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2 years ago

Read 2 more answers

Need help with the introduction to probability lesson, photo is attached below.

erastovalidia [21]

Answer:

For A.) there are only 2 so 2/8 would become0.25

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Step-by-step explanation:

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8 months ago