Let, one leg be x units.
since the other leg is 9.0 units shorter than the other,
the measure of the other leg =x-9
hypotenuse =13.0 units
According to Pythagorean theorem, the square of the hypotenuse is equal to the sum of squares of the other two sides.
x^2 +(x-9)^2 =13^2
x^2 + x^2 -18x +81 = 169
2x^2-18x-88=0
Divide the whole equation by 2
x^2 - 9x -44 =0
Let's use the quadratic formula to find the roots.
formula: x= [-b ± sqrt(b^2-4*a*c)]/2*a
a=1 b=-9 c=-44
x= [9±sqrt(81+176)]/2
=[9±sqrt(257)]/2
=[9±16.03]/2
=25.03/2 or -7.03/2
length of a triangle cannot be negative,
Therefore, x=25.03/2 =12.52
one leg =12.5 units
other leg =12.52-9 =3.5 units
Answer:
c) 8.6
Step-by-step explanation:
Take the coordinates of the two points.
(0,8) and (7,3)
Formula:
√(x₁ - x₂)² + (y₁ - y₂)²
Substitute:
√(0 - 7)² + (8 - 3)²
√(-7)² + (5)²
√(49) + (25)
√(74) ≈ 8.6
Answer:
True
Step-by-step explanation:
-11 is greater than -16
X = -m/b.
Subtract z from both side and you get
-m= bx
Now divide both side by b and you get
-m/b= x
Part A: f(t) = t² + 6t - 20
u = t² + 6t - 20
+ 20 + 20
u + 20 = t² + 6t
u + 20 + 9 = t² + 6t + 9
u + 29 = t² + 3t + 3t + 9
u + 29 = t(t) + t(3) + 3(t) + 3(3)
u + 29 = t(t + 3) + 3(t + 3)
u + 29 = (t + 3)(t + 3)
u + 29 = (t + 3)²
- 29 - 29
u = (t + 3)² - 29
Part B: The vertex is (-3, -29). The graph shows that it is a minimum because it shows that there is a positive sign before the x²-term, making the parabola open up and has a minimum vertex of (-3, -29).
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Part A: g(t) = 48.8t + 28 h(t) = -16t² + 90t + 50
| t | g(t) | | t | h(t) |
|-4|-167.2| | -4 | -566 |
|-3|-118.4| | -3 | -364 |
|-2| -69.6 | | -2 | -194 |
|-1| -20.8 | | -1 | -56 |
|0 | -28 | | 0 | 50 |
|1 | 76.8 | | 1 | 124 |
|2 | 125.6| | 2 | 166 |
|3 | 174.4| | 3 | 176 |
|4 | 223.2| | 4 | 154 |
The two seconds that the solution of g(t) and h(t) is located is between -1 and 4 seconds because it shows that they have two solutions, making it between -1 and 4 seconds.
Part B: The solution from Part A means that you have to find two solutions in order to know where the solutions of the two functions are located at.
