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valentinak56 [21]
3 years ago
14

Please help asap will mark brainliest

Mathematics
2 answers:
Keith_Richards [23]3 years ago
7 0

Answer: The first one. -6x^2+7x

Step-by-step explanation:

The -x when distributed to the terms in the parentheses, flips their signs.

steposvetlana [31]3 years ago
6 0
The answer is -6x^2 + 7x, or answer choice A.
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List L consists of numbers 1, square root of 2, x, and x^2 , where x>0, and the range of the numbers in list L is 4. Which qu
Hitman42 [59]
If the range of the numbers in the list L is 4, then x can't be 4 since there is an element in the list that is x^2. Therefore,
0 ≤ x^2 ≤ 4 so that it would be the highest number and
0 ≤x ≤ 2

So, the value of x can be equal to 2 but not higher. So, 2 is always greater or equal to x.
3 0
3 years ago
NO LINKS!!!!
statuscvo [17]

The instructions for these problems seem incomplete. I'm assuming your teacher wants you to find the equation of each parabola.

===============================================================

Problem 1

Let's place Maya at the origin (0,0) on the xy coordinate grid. We'll have her kick to the right along the positive x axis direction.

The ball lands 40 feet away from her after it sails through the air. So the ball lands at (40,0). At the halfway point is the vertex (due to symmetry of the parabola), so it occurs when x = 40/2 = 20. The ball is at a height of 18 feet here, which means the vertex location is (20,18).

The vertex being (h,k) = (20,18) leads to...

y = a(x-h)^2 + k\\y = a(x-20)^2 + 18

Let's plug in another point on this parabola, say the origin point. Then we'll solve for the variable 'a'.

y=a(x-20)^2+18\\\\0 = a(0-20)^2 + 18\\\\0 = a(-20)^2 + 18\\\\0 = 400a + 18\\\\-18 = 400a\\\\400a = -18\\\\a = -\frac{18}{400}\\\\a = -\frac{9}{200}

So we can then say,

y = a(x-h)^2 + k\\\\y = -\frac{9}{200}(x-20)^2 + 18\\\\y = -\frac{9}{200}(x^2-40x+400) + 18\\\\y = -\frac{9}{200}x^2-\frac{9}{200}*(-40x)-\frac{9}{200}*400 + 18\\\\y = -\frac{9}{200}x^2+\frac{9}{5}x-18 + 18\\\\y = -\frac{9}{200}x^2+\frac{9}{5}x\\\\

The final equation is in the form y = ax^2+bx+c where a = -\frac{9}{200}, \ b = \frac{9}{5}, \text{ and } c = 0

x = horizontal distance the ball is from Maya

y = vertical distance the ball is from Maya

Maya is placed at the origin (0,0)

The graph is shown below. Refer to the blue curve.

===============================================================

Problem 2

We could follow the same steps as problem 1, but I'll take a different approach.

Like before, the kicker is placed at the origin and will aim to the right.

Since the ball is on the ground at (0,0), this is one of the x intercepts. The other x intercept is at (60,0) because it lands 60 feet away from the kicker.

The two roots x = 0 and x = 60 lead to the factors x and x-60 respectively.

We then end up with the factorized form y = ax(x-60) where the 'a' is in the same role as before. It's the leading coefficient.

To find 'a', we'll plug in the coordinates of the vertex point (30,20). The 30 is due to it being the midpoint of x = 0 and x = 60. The 20 being the height of the ball at this peak.

y = ax(x-60)\\\\20 = a*30(30-60)\\\\20 = a*30(-30)\\\\20 = -900a\\\\a = -\frac{20}{900}\\\\a = -\frac{1}{45}

Let's use this to find the standard form of the parabola.

y = ax(x-60)\\\\y = -\frac{1}{45}x(x-60)\\\\y = -\frac{1}{45}(x^2-60x)\\\\y = -\frac{1}{45}*x^2-\frac{1}{45}*(-60x)\\\\y = -\frac{1}{45}x^2+\frac{4}{3}x\\\\

Refer to the red curve in the graph below.

===============================================================

Problem 3

We can use either method (similar to problem 1 or problem 2). The second problem's method is probably faster.

Logan is placed at (0,0) and kicks to the right. The ball lands at (30,0). Those x intercepts are x = 0 and x = 30 respectively, which lead to the factors x and x-30. This leads to y = ax(x-30)\\\\

The midpoint of (0,0) and (30,0) is (15,0). Eight feet above this midpoint is the location (15,8) which is the vertex. Plug in (x,y) = (15,8) and solve for 'a'

y = ax(x-30)\\\\8 = a*15(15-30)\\\\8 = a*15(-15)\\\\8 = -225a\\\\a = -\frac{8}{225}\\\\

So,

y = ax(x-30)\\\\y = -\frac{8}{225}x(x-30)\\\\y = -\frac{8}{225}(x^2-30x)\\\\y = -\frac{8}{225}*x^2-\frac{8}{225}*(-30x)\\\\y = -\frac{8}{225}x^2+\frac{16}{15}x\\\\

The graph is the green curve in the diagram below.

Like with the others, x and y represent the horizontal and vertical distance the ball is from the kicker. The kicker is placed at the origin (0,0).

Once we know the equation of the parabola, we can answer questions like: "how high up is the ball when it is horizontally 10 feet away?". We do this by plugging in x = 10 and computing y.

Side note: We assume that there isn't any wind. Otherwise, the wind would slow the ball down and it wouldn't be a true parabola. However, that greatly complicates the problem.

5 0
2 years ago
Point T is on line segment \overline{SU} SU . Given ST=2x+6,ST=2x+6, TU=4,TU=4, and SU=4x,SU=4x, determine the numerical length
vazorg [7]

Answer: The numerical length of  \overline{SU} is 20 units.

Step-by-step explanation:

Given: Point T is on line segment \overline{SU} .

i.e. \overline{SU}=\overline{ST}+\overline{TU}             (i)

Since , it is given that ST=2x+6,TU=4, and SU=4x

Substitute these values in (i), we get

4x= 2x+6 + 4\\\\\Rightarrow\ 4x-2x=10\\\\\Rightarrow\ 2x=10\\\\\Rightarrow\ x=5

Now, the numerical length of \overline{SU}= 4(5)=20\ units

Hence, the  numerical length of  \overline{SU} is 20 units.

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