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myrzilka [38]
2 years ago
5

Which function rule best models the data in the table. X: 0,1,2,3 y: 3,4,7,12

Mathematics
1 answer:
Lisa [10]2 years ago
4 0

Answer:

x = 3y + 4x

its how graphs work

You might be interested in
A triangle has sides 12, 35, and 37. Identify whether the triangle is right, acute, or obtuse.
s2008m [1.1K]

Answer:

Right triangle

Step-by-step explanation:

Using pythagorean theorem,

c^2 = a^2 + b^2\\37^2 = 12^2 + 35^2 \\1369 = 144 + 1225\\1369 = 1369

Hence satisfied. So it is right triangle.

6 0
3 years ago
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 want the answer of question 27 with explanation
m_a_m_a [10]

Answer:

6√2

Step-by-step explanation:

√32+√18-1√4

√16×2+√9×2-1√2

4√2+3√2-1√2

6√2

7 0
3 years ago
How do you complete the square to rewrite y= x^ -6x + 15 in vertex form
Vaselesa [24]
Vertex form of a quadratic equation is y=a(x-h)^{2}+k
where (h, k) is the vertex. 

Our aim is to write the expression y=x^{2} -6x+15 in the form 
y=a(x-h)^{2}+k



y=x^{2} -6x+15

y=x^{2} -2*3*x+15

y=(x^{2} -2*3*x+ 3^{2})-3^{2}+ 15

y=(x-3)^{2}-9+15

y=(x-3)^{2}+6
5 0
3 years ago
Solve: -2x - 5 = -3 <br> plz help me hurry
lord [1]

Answer:

x=-1

Step-by-step explanation:

-2x - 5 = -3

add 5 to both sides

-2x = 2

divide both sides by -2

x = -1

8 0
2 years ago
Read 2 more answers
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