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3 years ago

X+y=14

Mathematics

1 answer:

RoseWind [281]3 years ago

5 0

Answer:

x = -18

y = 4

Step-by-step explanation:

x + y = 4

2x + 3y = 30

You can do any math operation to each side of the equal sign as long as you do the same operation to both sides.

change x + y = 4 to x = 4 - y

Now substitute (4 - y) into the second equation, in place of x.

2(4-y) + 3y = 30

multiply the 2(4-y) part:

8 -2y + 3y = 30

simplify:

8 + y = 30

subtract 8 from each side:

y = 22

now substitute the value of y back into the first equation to solve for x:

x + 22 = 4

subtract 22 from both sides:

x = -18

Check:

x + y = 4

22 -18 = 4

4 = 4 (this checks)

2(-18) + 3(22) = 30

-36 + 66 = 30

30 =30 (this checks also)

PS: you could have solved for y first if you had wanted to. The results would be the same.

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Answer:

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Step-by-step explanation:

a) The function in terms of time and the inital angle measured from the horizontal is:

$\vec r (t) = [(v_{o}\cdot \cos \theta)\cdot t]\cdot i + \left[(v_{o}\cdot \sin \theta)\cdot t -\frac{1}{2}\cdot g \cdot t^{2} \right]\cdot j$

The particular expression for the cannonball is:

$\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$

b) The components of the position of the cannonball before hitting the ground is:

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After a quick substitution and some algebraic and trigonometric handling, the following expression is found:

$0 = 90\cdot \sin \theta - 6\cdot \left(\frac{x}{90\cdot \cos \theta} \right)$

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$6\cdot x = 4050\cdot \sin 2\theta$

$x = 675\cdot \sin 2\theta$

The angle for a maximum horizontal distance is determined by deriving the function, equalizing the resulting formula to zero and finding the angle:

$\frac{dx}{d\theta} = 1350\cdot \cos 2\theta$

$1350\cdot \cos 2\theta = 0$

$\cos 2\theta = 0$

$2\theta = \frac{\pi}{2}$

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Now, it is required to demonstrate that critical point leads to a maximum. The second derivative is:

$\frac{d^{2}x}{d\theta^{2}} = -2700\cdot \sin 2\theta$

$\frac{d^{2}x}{d\theta^{2}} = -2700$

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$y = 45\cdot t - 6\cdot t^{2}$

The maximum height can be found by deriving the previous expression, which is equalized to zero and critical values are found afterwards:

$\frac{dy}{dt} = 45 - 12\cdot t$

$45-12\cdot t = 0$

$t = \frac{45}{12}$

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Now, the second derivative is used to check if such solution leads to a maximum:

$\frac{d^{2}y}{dt^{2}} = -12$

Which demonstrates the assumption.

The maximum height reached by the cannonball is:

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