we have that
using a graph tool
see the attached figure
statements
case a) The domain is {x|x ≤ –2}-------> Is False
the domain is all real numbers---------> the interval (-∞,∞)
case b) The range is {y|y ≤ 6}.------> Is True
The range is the interval (-∞,6]
case c) The function is increasing over the interval (–∞ , –2).-----> Is True
See the attached figure
case d)The function is decreasing over the interval (−4, ∞).-----> Is False
In the interval (-4,-2) the function is increasing and in the interval (-2,∞) the function is decreasing (See the attached figure)
case e)The function has a positive y-intercept.------> Is True
The value of y-intercept is 
Answer:
Hola
Step-by-step explanation:
Well if this is based off the first thing I answered, then 1. C
as I told you before, every time the x axis values go up one, the y values go up by x2 (aka times 2). Therefore it’s C.
2. It is linear because the unit change (amount it goes up aka slope) is the same since it’s alway x2. Hence the line on a graph would look linear
Answer:
Step-by-step explanation:
The question is incomplete. Here is the complete question
On a flight New York to London an airplane travels at a constant speed. An equation relating the distance traveled in miles d to the number of hours flying t is t= 1/500d. How long will it take the airplane to travel 800 miles?
Speed is the change in distance if a body with respect to time. It is expressed mathematically according to the question as t = 1/500 × d
To determine the time it will take to travel 800miles, we will substitute d = 800 into the modeled equation.
t = 1/500 × 800
t = 800/500
t = 8/5
t = 1.6
Answer:
False
Step-by-step explanation:
the area (A) of a rectangle is calculated as
A = length × breadth
= (x + 5)(x - 5) ← expand factors using FOIL
= x² - 5x + 5x - 25
= x² - 25 ≠ x² - x - 15
h(t)=(t+3) 2 +5 h, left parenthesis, t, right parenthesis, equals, left parenthesis, t, plus, 3, right parenthesis, squared, plu
lesya692 [45]
Answer:
1
Step-by-step explanation:
If I understand the question right, G(t) = -((t-1)^2) + 5 and we want to solve for the average rate of change over the interval −4 ≤ t ≤ 5.
A function for the rate of change of G(t) is given by G'(t).
G'(t) = d/dt(-((t-1)^2) + 5). We solve this by using the chain rule.
d/dt(-((t-1)^2) + 5) = d/dt(-((t-1)^2)) + d/dt(5) = -2(t-1)*d/dt(t-`1) + 0 = (-2t + 2)*1 = -2t + 2
G'(t) = -2t + 2
This is a linear equation, and the average value of a linear equation f(x) over a range can be found by (f(min) + f(max))/2.
So the average value of G'(t) over −4 ≤ t ≤ 5 is given by ((-2(-4) + 2) + (-2(5) + 2))/2 = ((8 + 2) + (-10 + 2))/2 = (10 - 8)/2 = 2/2 = 1
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