Answer:
The factors of x² - 3·x - 18, are;
(x - 6), (x + 3)
Step-by-step explanation:
The given quadratic expression is presented as follows;
x² - 3·x - 18
To factorize the given expression, we look for two numbers, which are the constant terms in the factors, such that the sum of the numbers is -3, while the product of the numbers is -18
By examination, we have the numbers -6, and 3, which gives;
-6 + 3 = -3
-6 × 3 = -18
Therefore, we can write;
x² - 3·x - 18 = (x - 6) × (x + 3)
Which gives;
(x - 6) × (x + 3) = x² + 3·x - 6·x - 18 = x² - 3·x - 18
Therefore, the factors of the expression, x² - 3·x - 18, are (x - 6) and (x + 3)
Answer: 275d>1650
Step-by-step explanation:
because the question ask you a distance at least 1650 which means 275 times the days has to be greater than 1650.
Have a nice day.
P.S if this helped plz consider making me brainliest
Answer:
5000
Step-by-step explanation:
Answer:
(11, 13)
Step-by-step explanation:
Carolyn's work is incorrect. Below is the correct solution:
Given:
----› Equation 1
-----› Equation 2
Substitute y = (x + 2) in equation 1
----› Equation 1
(substitution)
(distributive property)
[Note: this is where Carolyn made a mistake]

Collect like terms

(addition property of equality).

Substitute x = 11 in equation 2
-----› Equation 2
(substitution)
✅The solution to the system of equations would be:
(11, 13)
Answer:
a.) Between 0.5 and 3 seconds.
Step-by-step explanation:
So I just went ahead and graphed this quadratic on Desmos so you could have an idea of what this looks like. A negative quadratic, and we're trying to find when the graph's y-values are greater than 26.
If you look at the graph, you can easily see that the quadratic crosses y = 26 at x-values 0.5 and 3. And, you can see that the quadratic's graph is actually above y = 26 between these two values, 0.5 and 3.
Because we know that the quadratic's graph models the projectile's motion, we can conclude that the projectile will also be above 26 feet between 0.5 and 3 seconds.
So, the answer is a.) between 0.5 and 3 seconds.