Calculate the slope of the line that passes through the following points:

Find the value of X and Y for all three questions. Show your work. Will give brainliest!

vodka [1.7K]

Answer:

(13)

x= 8

y=11

(14)

x= 20

y=20.75

(15)

x=9

y=13

Step-by-step explanation:

(13) 8x-1 = 11x -25

24=3x

x= 8

8(8)-1 +15y-48= 180

15y=165

y=11

(14) 7x-44 + 4x + 4= 180

11x= 220

x= 20

4(20)+4+39= 8y-43

123=8y-43

166=8y

y=20.75

(15) 12x+1=15x-26

3x=27

x=9

28+15(9)-26+4y-9=180

4y=52

y=13

3 0

3 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:

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

$y_{max} = 84.375\,m$

7 0

3 years ago

I NEED HELP PLEASE, THANKS!

Finger [1]

Answer:

the 3rd option is the answer

Step-by-step explanation:

I hope the attached file is self-explanatory

3 0

3 years ago

Select ALL the correct answers.

jolli1 [7]

Answer:

(4x+5)(-3x-1) = -12x²-19x-5

Option A:

(-16x² + 10x - 3) + (4x² - 29x - 2) = -12x²-19x-5

Option A is correct.

Option B:

3(x - 5) - 2(6x² + 9x + 5) = -12x²-15x-25

Option B is wrong.

Option C:

2(x - 1) - 3(4x² + 7x + 1) = -12x²-19x-5

Option C is correct.

Option D:

(2x² - 11x - 9) - (14x² + 8x - 4) = -12x²-19x-5

Option D is correct.

4 0

2 years ago

I have a bunch of graph problems and they are killin me! Please answer 90 pnts, giving brainliest

stiv31 [10]

Answer:

first picture d

second d

third a

forth a

5th c

im not that great at math tho

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers