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Ulleksa [173]
3 years ago
6

write a linear equation that intersects y=x^2 at two points. Then write a second linear equation that intersects y=x^2 at just o

ne point, and a third linear that does not intersect y=x^2. Explain how you found the linear equations.
Mathematics
1 answer:
Liula [17]3 years ago
5 0

We know that y = x^2 is a parabola, concave up, with vertex in the origin (0,0)

So, we may use three horizontal lines for our purpose: any horizontal line above the x axis will intersect the parabola twice. The x axis itself intersects the parabola once on the vertex, while any horizontal line below the x axis won't intercept the parabola.

Here's the examples:

  • The horizontal line y = 4 intercepts the parabola twice: the system y = x^2,\ y = 4 is solved by x^2=4 \implies x = \pm 2
  • The horizontal line y=0 intercepts the parabola only once: the system is y=x^2,\ y=0 which yields x^2=0\implies x=0 which is a repeated solution
  • The horizontal line y=-5 intercepts the parabola only once: the system is y=x^2,\ y=-5 which yields x^2=-5 which is impossible, because a squared number can't be negative.
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A ball is thrown vertically in the air with a velocity of 95 ft/s. The ball is at a height of 120 ft.
sattari [20]

Answer:

The ball is at a height of 120 feet after 1.8 and 4.1 seconds.

Step-by-step explanation:

The statement is incomplete. The complete statement is perhaps the following "A ball is thrown vertically in the air with a velocity of 95 ft/s. What time in seconds is the ball at a height of 120ft. Round to the nearest tenth of a second."

Since the ball is launched upwards, gravity decelerates it up to rest and moves downwards. The position of the ball can be determined as a function of time by using this expression:

y = y_{o} + v_{o}\cdot t +\frac{1}{2}\cdot g \cdot t^{2}

Where:

y_{o} - Initial height of the ball, measured in feet.

v_{o} - Initial speed of the ball, measured in feet per second.

g - Gravitational constant, equal to -32.174\,\frac{ft}{s^{2}}.

t - Time, measured in seconds.

Given that y_{o} = 0\,ft, v_{o} = 95\,\frac{ft}{s}, g = -32.174\,\frac{ft}{s^{2}} and y = 120\,ft, the following second-order polynomial is found:

120\,ft = 0\,ft + \left(95\,\frac{ft}{s} \right)\cdot t +\frac{1}{2}\cdot \left(-32.174\,\frac{ft}{s^{2}} \right) \cdot t^{2}

-16.087\cdot t^{2} + 95\cdot t -120 =0

The roots of this polynomial are, respectively:

t_{1} \approx 4.075\,s and t_{2} \approx 1.831\,s.

Both roots solutions are physically reasonable, since t_{1} represents the instant when the ball reaches a height of 120 ft before reaching maximum height, whereas t_{2} represents the instant when the ball the same height after reaching maximum height.

In nutshell, the ball is at a height of 120 feet after 1.8 and 4.1 seconds.

8 0
3 years ago
Consider a single spin of the spinner.
Nitella [24]

Answer:

landing on a shaded portion and landing on an even number ; landing on a shaded portion and landing on a number greater than 3 ; landing on an unshaded portion and landing on an odd number ; and landing on an unshaded portion and landing on a number less than 2.

Explanation:

Mutually exclusive events are events that cannot occur at the same time.  Our spinner has two grey sections, on 1 and 3; and two white sections, on 2 and 4.

This means that the spinner cannot land on a shaded (grey) portion and land on an even number at the same time, since the grey sections are numbered 2 and 4, both of which are even numbers.

The spinner also cannot land on a shaded (grey) portion and land on a number greater than 3 at the same time; this is because the only number greater than 3 on the spinner is 4, which is a white portion.

The spinner cannot land on an unshaded (white) portion and land on an odd number, since the white sections are labeled 2 and 4, which are both even.

The only number on the spinner less than 2 is 1, which is grey; this means the spinner cannot land on a number less than 2 and an unshaded (white) portion at the same time.

5 0
3 years ago
Read 2 more answers
Peter drove 1.5 hours at an average speed of 40 mph. He then drove for another 40km.
egoroff_w [7]

Answer:

john drove for 3 hours at a rate of 50 miles per hour and for 2 hours at 60 miles per hour. 

Step-by-step explanation:

5 0
2 years ago
Quadrilateral OPQR is inscribed inside a circle as shown below. Write a proof showing that angles O and Q are supplementary
k0ka [10]
Hello,

very simple they intercept supplementary arcs. (making a sum of 360°/2 for inscribed angles )
4 0
3 years ago
If n = 11 and p = 5 what is the value of the expression 18 - n - 2 + p
yan [13]

Answer:

10

Step-by-step explanation:

If n = 11 and p = 5, we can just substitute the values in for n and p in the equation 18 - n - 2 + p.


The new equation would be 18 - 11 - 2 + 5.

Using PEMDAS, we get the answer of 10.  

18 - 11 = 7

7 - 2 = 5

5 + 5 = 10


Hope this helps!

3 0
2 years ago
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