Help please I am very confused

Mathematics

1 answer:

Helga [31]3 years ago

3 0

The sides of the triangle occur in a ratio of 4 : 7 : 2, so if x is some positive number, then we can write each side's length in terms of x as 4x, 7x, and 2x.

The perimeter is 299 yd, so

4x + 7x + 2x = 299 yd

13x = 299 yd

x = (299 yd) / 13

x = 23 yd

Then the sides of the triangle have lengths of

4x = 4 • 23 yd = 104 yd

7x = 7 • 23 yd = 161 yd

2x = 2 • 23 yd = 46 yd

"Median" here refers to the side length between the shortest and longest sides, so the answer would be 104 yd.

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A firework is fired into the air from the top of a barn. Tis hight (h) above the ground in yards after (t) seconds is given by t

Grace [21]

The given function for the height of the firework is a quadratic function

1. Time at which the firework reaches the maximum height is 1 seconds

2. The maximum height of the firework, is 25 yards

3. Time after which the firework will fall to the ground is (1 + √5) seconds

Reason:

The given function that represents the height of the fireworks with time is

presented as follows;

h(t) = -5·t² + 10·t + 20

1. The time at which the firework reaches its maximum height is given by

the maximum point of the given function as follows;

The x-value of the maximum point of a quadratic function is $x = \dfrac{b}{2 \cdot a}$

Where;

a, and b, are the coefficient of x² and x, in the general form of a quadratic function f(x) = a·x² + b·x + c

By comparison, we have;

$t = -\dfrac{10}{2 \times (-5)} = 1$

The time at which the firework reaches the maximum height is t = 1 seconds

2. The maximum height is given by plugging in the value of t, at the maximum point into the given function as follows;

h(1) = -5×1² + 10×1 + 20 = 25

The maximum height of the firework, f(1) = 25 yards

3 The time at which the firework will fall to the ground, is given by the zero of the function as follows;

When the firework falls to the ground, h(t) = 0 = -5·t² + 10·t + 20

Dividing both sides by (-5) gives;

$\dfrac{0}{-5} = \dfrac{ -5 \cdot t^2 + 10 \cdot t + 20}{-5} = t^2 - 2 \cdot t - 4$

t² - 2·t - 4 = 0

By the quadratic formula $x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}$ , we get;

$t = \dfrac{2\pm \sqrt{(-2)^{2}-4\times 1\times (-4)}}{2\times 1} = \dfrac{2\pm \sqrt{20}}{2\times 1} = \dfrac{2\pm 2 \times \sqrt{5}}{2\times 1} = 1 \pm \sqrt{5}$