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Sunny_sXe [5.5K]

3 years ago

13

Can someone answer this question correctly it's my test please don't answer if you don't understand please show work I need it t

oday thank you

1 answer:

saveliy_v [14]3 years ago

6 0

Answer:

The answer is b.) -5.2 degrees

Step-by-step explanation:

to find the mean of this problem you have to add all numbers and then divide it by how many numbers there is.

so you have to add -42+ -17+14+-4+23 and that'll equal -26

so you take -26 and divide it by 5 because thats how many numbers their are to divide

-26 divided by 5 is (-5.2)

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What is the standard form equation of the line shown below? Graph of a line going through negative 1, 5 and 2, 4

natima [27]

Answer:

y = -1/3x + 14/3

Step-by-step explanation:

We have the standard equation as;

y = mx + b

where m is the slope and b is the y-intercept

the given points are;

(-1,5) and (2,4)

So we plug x and y for each of the points

then we create two linear equations which we can solve simultaneously to get the values of m and b

From the points;

5 = -m + b•••••••(i)

4 = 2m + b

from i;

b = 5 + m

substitute this into equation ii

4 = 2m + 5 + m

4 = 3m + 5

4-5 = 3m

3m = -1

m = -1/3

Recall;

b = 5 + m

b = 5-1/3

b = 14/3

so we have the equation as;

y = -1/3x + 14/3

Multiply through by 3

3y = -x + 14

7 0

3 years ago

The table of values shown below represents a linear function. Which of these points could also be an ordered pair in the table,

bearhunter [10]

Answer:

(b) (18, 27), because the rate of change of the function is 4/3

hope this helps!

Step-by-step explanation:

6 0

2 years ago

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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:

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

$y_{max} = 84.375\,m$

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3 years ago

72.1 written as a fraction

kaheart [24]

It is 72 1/10. This is because 72 is a whole number and point one would be tenths.

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3 years ago

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RoseWind [281]

I’m pretty sure it’s x-3y over 2

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3 years ago