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BigorU [14]
3 years ago
12

Math graph and solution y=3x2+18x-21

Mathematics
1 answer:
Likurg_2 [28]3 years ago
6 0
We have that

<span>y=3x</span>²<span>+18x-21

using a graph tool
see the attached figure

the solutions are
x=-7
x=1</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}}

5 0
3 years ago
For breakfast, Mr hill bought a cup of coffee for 1.39 and a bagel for 1.85. How much change will he get for 5.00
notsponge [240]
Me hill will get 1.76 in change
3 0
3 years ago
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70% of the 30 sandcastles at Donatello Beach have moats. how many more moats are there​
bixtya [17]

Answer:

21 moats

Step-by-step explanation:

.70 x 30 = 21

4 0
3 years ago
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Two airplanes are flying in the air at the same height. Airplane A is flying east at 300mi/h and airplane B is flying north at 2
Ipatiy [6.2K]

Answer:

\frac{dAB}{dt}=340 mil/h

Step-by-step explanation:

The change of distance over time of the plain A is 300 mi/hour and 200 mi/hour for plane B. O is the point of the airport.

So, the distance from A to O AO = 90 miles and BO = 120 miles.

Now, we have a right triangle here.  We can use the Pythagorean theorem, so the distance between the planes will be:

AB^{2} =AO^{2}+BO^{2} (1)

AB =\sqrt{AO^{2}+BO^{2}}=

AB =\sqrt{90^{2}+120^{2}}=150 miles

If we take the derivative of the equation (1) we could find the change of the distance between planes.

2AB\frac{dAB}{dt}=2AO\frac{dAO}{dt}+2BO\frac{dBO}{dt}

2*150\frac{dAB}{dt}=2*90*300+2*120*200=102000 mil/h

\frac{dAB}{dt}=\frac{102000}{150*2}

Finally,

\frac{dAB}{dt}=340 mil/h

I hope it helps you!

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Answer:

Derek needs to add 3/4 cups of flour.

Step-by-step explanation:

If Derek has already used 1/4 cup of flour, he needs 3 more cups of flour to get to 1 full cup of flour.

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