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3 years ago

Hey yall so im taking a test and imm very confused on hwo to find the letter n for the equation

Mathematics

1 answer:

kotegsom [21]3 years ago

8 0

Answer:

n = -7,6

Step-by-step explanation:

n(n+1)+3=45

distributive property: a(b+c) = ab+ac

n²+n+3 = 45

n²+n -42 = 0

factor now

(n+7)(n-6) = 0

n = -7,6

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On an alien planet with no atmosphere, acceleration due to gravity is given by g = 12m/s^2. A cannonball is launched from the or

almond37 [142]

Answer:

a) $\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$ , b) $\theta = \frac{\pi}{4}$ , c) $y_{max} = 84.375\,m$ , $t = 3.75\,s$ .

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:

$x = (90\cdot \cos \theta)\cdot t$

$0 = 90\cdot \sin \theta - 6\cdot t$

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

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

$0 = 8100\cdot \sin \theta \cdot \cos \theta - 6\cdot x$

$0 = 4050\cdot \sin 2\theta - 6\cdot x$

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

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

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

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

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

$\cos 2\theta = 0$

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

$\theta = \frac{\pi}{4}$

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

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

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

Which demonstrates the existence of the maximum associated with the critical point found before.

c) The equation for the vertical component of position is:

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

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

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

$45-12\cdot t = 0$

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

$t = 3.75\,s$

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:

$y_{max} = 45\cdot (3.75\,s)-6\cdot (3.75\,s)^{2}$

$y_{max} = 84.375\,m$

7 0

3 years ago

GIVING AWAY BRAINLIEST READ THE QUESTION!

ozzi

Answer:

Step-by-step explanation:

A function is constant when the y value does not change when the x value does

The function is constant during section C

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3 years ago

Measuring and arcs.

n200080 [17]

Answer:

Hope this helps

Step-by-step explanation:

3. Minor arc, 44°

4. Major arc, 140°

5. Minor arc, 44°

6. Major arc, 316°

7. Semi circle, 180°

8. Major arc, 140°

9. 38°

10. 52°

11. 142°

12. 128°

13. 232°

14. 308°

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3 years ago

(2-4d)(2+4d)=0<br> Help please, Its the 0 property

lianna [129]

Answer:

d1=-1/2,d2=1/2

Step-by-step explanation:

hope this helps

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3 years ago

PLS HELP ANSWER ASAP MARK BRAINLEIST

Alenkinab [10]

X is always dependant to Y. so that should answer the 1st one. the 2nd looks to be around 17-18. 17.5 maybe. and the 3rd is yes cause he does not go to the pool so it isnt any cost.

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3 years ago