Question

Please help with these 3! Will mark brainliest. Thank you!

Answer 1

#1
The two functions are: $y=-8 x^{2} -2$ and $y=-8 x^{2}$ so the difference is that -2. Adding or subtracting a constant moves a function up (if the constant is positive) or down (if the constant is negative). So adding -2 means the function will shift down 2 units. That is, the last choice given.

#2
The horizontal distance the rocket travels is given by x. As the rocket is launched and travels a parabolic path up into the sky and back down it moves a certain distance (horizontally) away from where it started. The rocket stops traveling when it hits the ground (when the height y is equal to 0). Therefore, we are being asked to find x when y = 0. We are being asked to solve the equation: $y=-.06 x^{2} +9.6x+5.4$. To solve this we can use the quadratic formula.

The quadratic formula is: $x={ \frac{-b(plus minus) \sqrt {b^{2} -4ac}}{2a} }$. We need to determine the values of a, b and c from the equation we are trying to solve. a is the coefficient (number in front of) $x^{2}$, b is the coefficient of x and c is the constant (the number by itself). So in this problem we have:
a=-.06
b=9.6
c=5.4
We need to plug these values into the formula and simplify. Since we are rounding to the nearest hundredth (two decimal places) I would carry out your calculations to at least 4 decimal places and then round at the end. This reduces errors due to rounding. Here is how we use the formula:
$x={ \frac{-9.6(plus minus) \sqrt {9.6^{2} -4(-.06)(5.4)}}{2(-.06)} }$
$x= \frac{-9.6plusminus \sqrt{92.16+1.296} }{-.12}$
$x= \frac{-9.6plusminus 9.6672}{-.12}$
Before we continue we need to recognize that our answer, being a distance, has to be positive. So even though there are two solutions to the equation only the positive one is correct. We arrive at this answer as follows:
(-9.6-9.6672643)/-.12=160.56

c)
This part is done the same way as part b. Again, the horizontal distance the rocket travels is given by x. As the rocket is launched and travels a parabolic path up into the sky and back down it moves a certain distance (horizontally) away from where it started. The rocket stops traveling when it hits the ground (when the height y is equal to 0). Therefore, we are being asked to find x when y = 0. We are being asked to solve the equation: $y=-.02 x^{2} +.8x+37$. To solve this we can use the quadratic formula.

The quadratic formula is: $x={ \frac{-b(plus minus) \sqrt {b^{2} -4ac}}{2a} }$. We need to determine the values of a, b and c from the equation we are trying to solve. a is the coefficient (number in front of) $x^{2}$, b is the coefficient of x and c is the constant (the number by itself). So in this problem we have:
a=-.02
b=.8
c=37
We need to plug these values into the formula and simplify. Since we are rounding to the nearest hundredth (two decimal places) I would carry out your calculations to at least 4 decimal places and then round at the end. This reduces errors due to rounding. Here is how we use the formula:
$x={ \frac{-.8(plus minus) \sqrt {.8^{2} -4(-.02)(37)}}{2(-.02)} }$
$x= \frac{-.8plusminus \sqrt{.64+.2.96} }{-.04}$
$x= \frac{-.8plusminus 1.8973}{-.04}$
Before we continue we need to recognize that our answer, being a distance, has to be positive. So even though there are two solutions to the equation only the positive one is correct. We arrive at this answer as follows:
(-.8-1.8973)/-.04=67.43