Answer:
8a
Step-by-step explanation:
Given
5a
+ 3a
← factor out
from each term
=
(5a + 3a)
= 8a
So, how far is the car from where it was at t = 0 is 40 m
<h3>Velocity of the car</h3>
Since the location x of the car in meters is given by the function x = 30t - 5t² where t is in seconds, we need to find the time at which its velocity is 10 m/s in the negative direction by differentiating x with respect to t to find its velocity, v.
So, v = dx/dt
= d(30t - 5t²)/dt
= d30t/dt - d5t²/dt
= 30 - 10t
When v is 10 m/s in the negative direction, v = -10 m/s.
So, v = 30 - 10t
-10 = 30 - 10t
-10 - 30 = -10t
-40 = -10t
t = -40/-10
t = 4 s
<h3>The distance at 4 s when its velocity is -10 m/s</h3>
Since at t = 4 s, its velocity is -10 m/s and x = 30t - 5t² is the car's location. The car's distance from t = 0 after its velocity is -10 m/s is
x(4) - x(0) = 30(4) - 5(4)² - [30(0) - 5(0)²]
= 120 - 5(16) - [0 - 0]
= 120 - 80 - 0
= 40 m
So, how far is the car from where it was at t = 0 is 40 m
Learn more about distance of car here:
brainly.com/question/17097458
Answer:
a. y = 2+4.3x when x = -6
substituting value of x, we get
<u>y=2+4.3*-6=2-25.8=23.8</u>
b. y = (x-3)² when x = 9
substituting value of x, we get
<u>y</u><u>=</u><u>(</u><u>9</u><u>-</u><u>3</u><u>)</u><u>²</u><u>=</u><u>6</u><u>²</u><u>=</u><u>3</u><u>6</u>
c. y=x-2 when x = 3.5
substituting value of x, we get
<u>y</u><u>=</u><u>3</u><u>.</u><u>5</u><u>-</u><u>2</u><u>=</u><u>1</u><u>.</u><u>2</u>
d. y=5x-4 when x = - 2
substituting value of x, we get
<u>y</u><u>=</u><u>5</u><u>*</u><u>-</u><u>2</u><u>-</u><u>4</u><u>=</u><u>-</u><u>1</u><u>0</u><u>-</u><u>4</u><u>=</u><u>-</u><u>1</u><u>4</u>