∠LQK+∠GQL=90º
Therefore:
(4n-15)+(3n)=90
7n=90+15
7n=105
n=105/7
n=15
∠LQK=4n-15=4(15)-15=60-15=45
Answer: ∠LQK=45º
Step-by-step explanation:
Two angles are called suplementary when their measures add to 180 degrees.
D+F= 180
3x+5+2x= 180
5x+5=180
5x= 175
x= 35°
angle D = 3x+5= 3(35)+5= <u>110°</u>
Bob is diving
he is at a depth of 2 meters below sea level
sam goes down 4 times as deep
how deep is sam
depth is a negative thing because you are 2 below 0 or -2
then we mutiply -2 by 4 to get 4 times to get -8 meters below sea level
Problem
For a quadratic equation function that models the height above ground of a projectile, how do you determine the maximum height, y, and time, x , when the projectile reaches the ground
Solution
We know that the x coordinate of a quadratic function is given by:
Vx= -b/2a
And the y coordinate correspond to the maximum value of y.
Then the best options are C and D but the best option is:
D) The maximum height is a y coordinate of the vertex of the quadratic function, which occurs when x = -b/2a
The projectile reaches the ground when the height is zero. The time when this occurs is the x-intercept of the zero of the function that is farthest to the right.
Answer:
2 and 6, 1 and 5
Step-by-step explanation: