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JulsSmile [24]
3 years ago
7

35 points please help!!!!! :(

Mathematics
1 answer:
Allisa [31]3 years ago
6 0
1. m = (y2 - y1) / (x2 - x1)
2. negative
3. y = 2
4. 0 slope
5. line A has steepest slope
6. (5,5)(-5,-1)...slope = -6/-10 = 3/5
7. Line B has a slope of -5
8. (1989,3)(1999,5.5) = 2.50/10 = 0.25
9. not sure
10. -1/4 inches per month
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Help please with this question?
xxMikexx [17]
G(x+1)= \sqrt{(x+1)^2+4} \\ G(x+1)= \sqrt{x^2+2x+1+4} \\  G(x+1)= \sqrt{x^2+2x+5}
7 0
3 years ago
An object is launched at 29.4 meters per
Lerok [7]

Answer:

The reasonable domain for the scenario is option 'a';

a) [0, 7]  

Step-by-step explanation:

For the projectile motion of the object, we are given;

The speed at which the object is launched, v = 29.4 meters per second

The height of the platform from which the object is launched, h = 34.3 meter

The equation for the height of the object as a function of time 'x' is given as follows;

f(x) = -4.9·x² + 29.4·x + 34.3

The domain for the scenario, is given by the possible values of 'x' for the function, which is found as follows;

At the height from which the object is launched, x = 0, and f(x) = 34.3

At the ground level to which the object can drop, f(x) = 0

∴ f(x) = -4.9·x² + 29.4·x + 34.3 = 0

-4.9·x² + 29.4·x + 34.3 = 0

By the quadratic formula, we have;

x = (-29.4 ± √(29.4² - 4 × (-4.9) × 34.3))/(2 × (-4.9)

∴ x = -1, or 7

Given that time is a natural number, we have the reasonable domain for the scenario as the start time when the object is launched, t = 0 to the time the object reaches the ground, t = 7

Therefore, the reasonable domain for the scenario is; 0 ≤ x ≤ 7 or [0, 7].

4 0
3 years ago
On Monday, the temperature at 10 a.m. at Sam’s house was –6° Fahrenheit. The
Alika [10]

Answer:

8

Step-by-step explanation:

to find the answer, subtract the second amount (2) from the first amount (-6), -6 - 2 = -8, the temperat decreased by 8

6 0
2 years ago
Read 2 more answers
Based on a study of population projections for 2000 to​ 2050, the projected population of a group of people​ (in millions) can b
Olegator [25]

Answer:

(a) 0.107 million per year

(b) 0.114 million per year

Step-by-step explanation:

A(t) = 11.19(1.009)^t

(a) The average rate of change between 2000 and 2014 is determined by dividing the difference in the populations in the two years by the number of years. In the year 2000, t=0 and in 2014, t=14. Mathematically,

\text{Rate}=\dfrac{A(2014) - A(2000)}{2014-2000}=\dfrac{11.19(1.009)^14-11.19(1.009)^0} {14}

A(0)=\dfrac{12.69-11.19}{14}=\dfrac{1.5}{14}=0.107

(b) The instantaneous rate of change is determined by finding the differential derivative at that year.

The result of differentiating functions of the firm y=a^x (where a is a constant) is \dfrac{dy}{dx}=a^x\ln a. Let's use in this in finding the derivative of A(t).

A\prime(t) = \dfrac{A}{t}=11.19\cdot1.009^t\ln1.009

In the year 2014, t=14.

A\prime(14) =11.19\cdot1.009^14\ln1.009=0.114

6 0
3 years ago
Can you help me out with this question 5 pls
Maslowich
Gives an output of 1 I think
3 0
3 years ago
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