1.A projectile's height (in metres) is given by the expression t(14-t), where time t is in seconds. How long is the projectile above a height of 40m?
2. The area of a right-angled triangle is 60 square units and the lengths of the two shorter sides differ by 7 units. Find the length of the hypotenuse.
the height is at 40 when
t^2-14t+40 = 0
(t-4)(t-10) = 0
SO, it is above 40 for 6 seconds (between t=4 and t=10)
x(x+7)/2 = 60
x^2+7x-120 = 0
(x+15)(x-8)=0
The triangle is the familiar 8-15-17 right triangle.
1. To find the length of time the projectile is above a height of 40m, we need to set up an equation and solve for t.
The given expression for the height of the projectile is t(14-t). We want to find the values of t for which the height is above 40m, so we can set up the inequality:
t(14-t) > 40
First, let's expand and rearrange the expression:
14t - t^2 > 40
Now, let's move all terms to one side and rewrite the inequality:
t^2 - 14t + 40 < 0
Next, we can solve this quadratic inequality by factoring or using the quadratic formula. Factoring is a bit simpler in this case:
(t - 4)(t - 10) < 0
To determine the values of t that satisfy the inequality, we need to consider the signs of the factors (t - 4) and (t - 10). The inequality will be true when one factor is positive and the other is negative.
We can create a sign chart:
| t < 4 | 4 < t < 10 | t > 10
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(t - 4) | positive | negative | negative
(t - 10) | positive | negative | positive
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Inequality| less than zero | less than zero | less than zero
From the sign chart, we can conclude that the inequality is satisfied when 4 < t < 10. Therefore, the projectile is above a height of 40m for a duration of 4 seconds < t < 10 seconds.
2. To find the length of the hypotenuse of a right-angled triangle, we can use the Pythagorean theorem. The theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Let's assume the lengths of the two shorter sides are a and b, and the length of the hypotenuse is c. We are given that the area of the triangle is 60 square units, so we can set up the following equation involving the product of the two shorter sides:
(a * b) / 2 = 60
Simplifying, we get:
ab = 120
We are also given that the lengths of the two shorter sides differ by 7 units, which can be written as:
b = a + 7
Now we can substitute this value into the previous equation:
a(a + 7) = 120
Expanding and rearranging the equation:
a^2 + 7a - 120 = 0
Now we can solve this quadratic equation by factoring or using the quadratic formula. Factoring gives us:
(a - 8)(a + 15) = 0
To find the lengths of the two sides, we can set each factor equal to zero:
a - 8 = 0 or a + 15 = 0
Solving these equations, we get:
a = 8 or a = -15
Since the length of a side cannot be negative, we discard the solution a = -15. Therefore, the length of one side is a = 8 units.
Using the equation b = a + 7, we can find the length of the other side:
b = 8 + 7 = 15 units
Now, we can find the length of the hypotenuse using the Pythagorean theorem:
c^2 = a^2 + b^2
c^2 = 8^2 + 15^2
c^2 = 64 + 225
c^2 = 289
Taking the square root of both sides to find c:
c = √289
c = 17 units
Therefore, the length of the hypotenuse is 17 units.