Step-by-step explanation:
making x subject in eq 1
x= (-1-9y)/-14
substituting the value of x in eq 2
17(-1-9y/-14)=7+9y
(-17-153y)/-14=7+9y
-17-153y=-98-126y
-17+98=153y-126y
81=27y
y=3
substituting this value of y in eq 1
-14x=-1-9(3)
-14x=-28
x=2
(x,y) = (2,3)
F(x) = 3x + 9
g(x) = 5x²
(f + g)(x) = (3x + 9) + 5x²
(f + g)(x) = 3x + 9 + 5x²
(f + g)(x) = 5x² + 3x + 9
The answer is probably like 8/11
The function is positive when x is larger than 3, then the interval in where the function is positive is: (3, ∞)
<h3>
on what interval is the function positive?</h3>
The function is positive when g(x) > 0.
Then we need to solve:
∛(x - 3) > 0
We can directly remove the cubic root, because it does not affect the sign of the argument, then:
x - 3 > 0
x > 3
The function is positive when x is larger than 3, then the interval in where the function is positive is: (3, ∞)
If you want to learn more about functions:
brainly.com/question/2456547
#SPJ1
285% is same as 285/100 which equal 2.85
Then 2.85 x 171 = 487.35
Answer: 487.35
