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3 years ago

Nsider the following system of equations y=6x^2+1, y=x^2+4

Mathematics

1 answer:

sp2606 [1]3 years ago

7 0

The given equations are y= 6x^2+1 and y=x^2+4

Both the equations equal y so we can make them equal

6x^2+1=x^2+4

To bring x term on one side we subtract x^2 both sides

6x^2-x^2+1=4

5x^2+1=4

To isolate x term we subtract 1 both sides

5x^2=4-1

5x^2=3

Dividing by 5 both sides

X^2=3/5

Taking root of both sides we have:

X=±√⅗

Substituting x value to find y

When x=+√⅗=0.7745

Y= 6x^2+1

Y=6(3/5)+1

Y= 18/5+1

Y=23/5 =4.6

When x= -√⅗=-0.7745

Y= 6(3/5)+1

Y=23/5=4.6

Point of intersections are ( 0.7745,4.6) and (-0.7745,4.6)

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So

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3 years ago

A surveyor leaves her base camp and drives 42km on a bearing of 032degree she then drives 28km on a bearing of 154degree,how far

ValentinkaMS [17]

Answer:

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

Step-by-step explanation:

The final position of the surveyor is represented by the following vectorial sum:

$\vec r = \vec r_{1} + \vec r_{2} + \vec r_{3}$ (1)

And this formula is expanded by definition of vectors in rectangular and polar form:

$(x,y) = r_{1}\cdot (\cos \theta_{1}, \sin \theta_{1}) + r_{2}\cdot (\cos \theta_{2}, \sin \theta_{2})$ (1b)

Where:

$x, y$ - Resulting coordinates of the final position of the surveyor with respect to origin, in kilometers.

$r_{1}, r_{2}$ - Length of each vector, in kilometers.

$\theta_{1}, \theta_{2}$ - Bearing of each vector in standard position, in sexagesimal degrees.

If we know that $r_{1} = 42\,km$ , $r_{2} = 28\,km$ , $\theta_{1} = 32^{\circ}$ and $\theta_{2} = 154^{\circ}$ , then the resulting coordinates of the final position of the surveyor is: