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3 years ago

How many hours are in 1/4+3/5

1 answer:

7 0

1/4 an hour is 60 divided by 4 times 1 which is 15 minutes.
3/5 an hour is 60 divided by 5 times 3 which is 36 minutes.
15 plus 36 equals 51 minutes which is less than 1 hour.

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A gardener is planting two types of trees: Type A is 9 feet tall and grows at a rate of 6 inches per year. Type B is 6 feet tall

xenn [34]

It will take exactly 3 years for these trees to be the same height

Step-by-step explanation:

A gardener is planting two types of trees

Type A is 9 feet tall and grows at a rate of 6 inches per year.
Type B is 6 feet tall and grows at a rate of 18 inches per year.

We need to find exactly how many years it will take for these trees to be the same height

To solve this problem let us make an equation of each type and then equate the two equations to find the number of years

Assume that the two types will have the same heights in x years

∵ Type A is 9 feet tall and grows at a rate of 6 inches per year

- Change the feet to inches

∵ 1 foot = 12 inches

∴ 9 feet = 9 × 12 = 108 inches

The height of the tree after x years is the sum of the initial height and the product of the rate of growth and the number of years

∵ The rate of growth = 6 inches per year

∵ The number of years = x

∵ The initial height = 108 inches

∴ The height of tree A = 108 + 6 x

∵ Type B is 6 feet tall and grows at a rate of 18 inches per year

∵ 1 foot = 12 inches

∴ 6 feet = 6 × 12 = 72 inches

∵ The rate of growth = 18 inches per year

∵ The number of years = x

∵ The initial height = 72 inches

∴ The height of tree B = 72 + 18 x

Now equate the two equations to find x

∵ 72 + 18 x = 108 + 6 x

- Subtract 72 from both sides

∴ 18 x = 36 + 6 x

- Subtract 6 x from both sides

∴ 12 x = 36

- Divide both sides by 12

∴ x = 3

It will take exactly 3 years for these trees to be the same height

Learn more:

You can learn more about the linear equations in brainly.com/question/9801816

#LearnwithBrainly

4 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.