1) [I don’t need any help with this question.]
2) [I don’t need any help with this question.]
3) A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y=-0.04x^2+8.3x+4.3, where x is the horizontal distance, in meters from the starting point on the roof and y is the height in meters, of the rocket above the ground. How far horizontally from its starting point will the rocket land? Round your answer to the nearest hundredth meter.
208.02 m<-----
416.03 m
0.52 m
208. 19 m
4) Landon is standing in a hole that is 6.5 m deep. he throws a rock, and it goes up into the air, out of the hole, and then lands on the ground above. The path of the rock can be modeled by the equation y=-0.05x^2+4.5x-6.5, where x is the horizontal distance of the rock, in meters, from landon and y is the height, in meters, of the rock above the ground. How far horizontally from landon will the rock land? Round your answer to the nearest hundredth of a meter. (1 point)
a. 82.03 m <----
b. 6.50 m
c. 90.00 m
d. 88.53 m
5) How many real number solutions does the equation have? (1 point)
0 = 5x^2 + 2x – 12
One solution
Two solutions
Infinitely many solutions
No solutions
6) How many real number solutions does the equation have? (1 point)
-8x^2 – 8x – 2 = 0
One solution
Two solutions
No solutions
Infinitely many solutions
7) Graph the set of points. (1 point)
(-6,4)(-4,0)(-3,-2)(-1,-7)
[I don't know the answer]
8) [I already know the answer to this one, so I don’t need help with this one.]
9) Find the solutions of the system.
y = x^2 + 3x - 4
y=2x + 2
(-3,6) and(2,-4)
(-3,-4) and (2,6)
(-3,-4) and (-2,-2) <----
No solution
10) Find the solutions to the system. (1 point)
y = x^2 + 8x + 2
11) If an object is dropped from a height of 85 feet, the function h(t) = - 16t^2 + 85 gives the height of the object after t seconds. Approximately, when will the object hit the ground? (1 point)
85.00 seconds
69.00 seconds
0.33 seconds
2.30 seconds
12) A ball is thrown into the air with an upward velocity of 32 feet per second. Its height, h, in feet after t
seconds is given by the function h(t) = –16t² + 32t + 6. What is the ball’s maximum height? How long does it
take the ball to reach its maximum height? Round to the nearest hundredth, if necessary. (1 point)
Reaches a maximum height of 22 feet after 1.00 second.
Reaches a maximum height of 22 feet after 2.00 seconds.
Reaches a maximum height of 44 feet after 2.17 seconds.
Reaches a maximum height of 11 feet after 2.17 seconds
First one is correct. The second one should be "d". You can check these answers by putting them into the original equation. You should get zero for an answer because "O" is when it hits the ground.
For 5 & 6 use the square root of b^2 - 4ac. If you get a zero under the root, there is one solution, if you get a positive number there are two and if you get a negative number there are two imaginary solutions.
7) make an x-axis and y-axis and plot the points (x,y) is the correct order.
2x+2 = x^2+3x-4
0 = x^2 +x -6
0 = (x-2)(x+3)
x = 2 x = -3
now find y but substituting.
I don't agree with the answer you have above.
10) set the quadratic equal to zero and then use the quadratic equation to solve.
x = -b +- square root of (b^2-4ac)
all of that is divide by 2a
11) the height has to be zero when it hits the ground. Set the equation = 0 and use the quadratic to solve. However, since this is MC, you could just substitute in the answers to see if you will get zero.
#12 recall that the vertex of a quadratic (max height in this case) occurs when
t = -b/2a = -32/-32 = 1
3) To find how far horizontally the rocket will land, we need to find the value of x when y=0. In other words, we need to find the value of x when the height of the rocket above the ground is zero. So we set the equation equal to zero:
-0.04x^2+8.3x+4.3 = 0
To solve this quadratic equation, we can use the quadratic formula. The quadratic formula states that for an equation of the form ax^2+bx+c=0, the solutions for x are given by:
x = (-b ± √(b^2-4ac)) / 2a
In our equation, a= -0.04, b= 8.3, and c= 4.3. Plugging in these values into the quadratic formula, we get:
x = (-8.3 ± √(8.3^2-4(-0.04)(4.3))) / 2(-0.04)
Simplifying this expression will give us two possible values of x. We can round the answers to the nearest hundredth meter to find how far horizontally the rocket will land.
4) Similar to the previous question, we need to find the value of x when y=0 in the equation -0.05x^2+4.5x-6.5 = 0. We can use the quadratic formula to solve this equation and find the values of x when y=0.
5) To find the number of real number solutions to the equation 0 = 5x^2 + 2x - 12, we can use the discriminant. The discriminant is the expression under the square root in the quadratic formula, given by b^2-4ac.
If the discriminant is positive, there are two real number solutions.
If the discriminant is zero, there is one real number solution.
If the discriminant is negative, there are no real number solutions.
In this case, we can calculate the discriminant and determine how many real number solutions exist for the equation.
6) Similar to the previous question, we need to use the discriminant to determine the number of real number solutions for the equation -8x^2 - 8x - 2 = 0.
7) To graph the set of points (-6,4), (-4,0), (-3,-2), and (-1,-7), we can plot each point on a graph using the x and y coordinates provided. Connecting the points will create a graphical representation of the set of points.
9) To find the solutions of the system of equations, we need to find the values of x and y that satisfy both equations simultaneously. One way to do this is by substitution or elimination. We can substitute the expression for y from one equation into the other equation and solve for x. Once we have the value of x, we can substitute it back into one of the equations to solve for y.
10) The instruction or equation to find the solutions to the system is missing. Please provide the complete equation or instructions.
11) To find when the object will hit the ground, we need to find the value of t when h(t)=0. This means we need to solve the equation -16t^2 + 85 = 0. We can use the quadratic formula to find the solutions for t, and then round the answer to the nearest hundredth second to determine approximately when the object will hit the ground.
12) To find the ball's maximum height, we need to find the vertex of the quadratic function h(t) = -16t² + 32t + 6. The vertex is given by the equation t = -b / 2a, where a, b, and c are the coefficients of the quadratic function. In this case, a=-16 and b=32. Plug these values into the equation to find the value of t at the maximum height. Then substitute this value of t back into the equation to find the maximum height reached by the ball in feet.