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dalvyx [7]
3 years ago
12

One model of Earth's population growth is P(t)= 64/(1+11e^0.8t)

Mathematics
1 answer:
Pepsi [2]3 years ago
5 0

Answer:

A. In 1991, there were 5.74 billion people  

D. The carrying capacity of Earth is 64 billion people

Have a nice day! :)

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EGF and GFH are a linear pair, m EFG = 4n + 15, and m GFH = 5n + 39. what are m EFG and m GFH?
Tasya [4]

The angles of ∠EFG  and ∠GFH are 71° and  109°

<h3>What are linear pair angles?</h3>

Linear pair of angles are formed when two lines intersect each other at a single point.

In other words, a linear pair of angles is a pair of adjacent angles formed when two lines intersect each other.

Linear pair angles are supplementary. This means the sum of  a linear pair angles is 180 degrees.

Therefore,

∠EFG + ∠GFH = 180

Therefore,

∠EFG = 4n + 15

∠GFH = 5n + 39

hence,

4n + 15 + 5n + 39 = 180

9n + 54 = 180

9n = 180 - 54

9n = 126

n = 126 / 9

n = 14

Hence,

∠EFG = 4n + 15 = 4(14) + 15 = 71°

∠GFH = 5n + 39 = 5(14) + 39 = 109°

learn more on linear pair angles here: brainly.com/question/28264317

#SPJ1

4 0
1 year ago
Please help me with a, b, and c
EastWind [94]
A: Deena subtracted 25, instead of adding it.
B: 9x - 25 = 88.
C: no idea
7 0
3 years ago
A worm travels 1250 millimeters how many meters did the worm travel
nataly862011 [7]
The worm traveled 1.25 meters
6 0
3 years ago
Read 2 more answers
Will mark brainest help please !!!
amm1812

8) red and 1, red and 2, red and 3, blue and 1, blue and 2, blue and 3

So answer is 6

9) 0.5*1/3 = 0.17

10) 0.5*2/3 = 0.33

7 0
3 years ago
PRECAL:<br> Having trouble on this review, need some help.
ra1l [238]

1. As you can tell from the function definition and plot, there's a discontinuity at x = -2. But in the limit from either side of x = -2, f(x) is approaching the value at the empty circle:

\displaystyle \lim_{x\to-2}f(x) = \lim_{x\to-2}(x-2) = -2-2 = \boxed{-4}

Basically, since x is approaching -2, we are talking about values of x such x ≠ 2. Then we can compute the limit by taking the expression from the definition of f(x) using that x ≠ 2.

2. f(x) is continuous at x = -1, so the limit can be computed directly again:

\displaystyle \lim_{x\to-1} f(x) = \lim_{x\to-1}(x-2) = -1-2=\boxed{-3}

3. Using the same reasoning as in (1), the limit would be the value of f(x) at the empty circle in the graph. So

\displaystyle \lim_{x\to-2}f(x) = \boxed{-1}

4. Your answer is correct; the limit doesn't exist because there is a jump discontinuity. f(x) approaches two different values depending on which direction x is approaching 2.

5. It's a bit difficult to see, but it looks like x is approaching 2 from above/from the right, in which case

\displaystyle \lim_{x\to2^+}f(x) = \boxed{0}

When x approaches 2 from above, we assume x > 2. And according to the plot, we have f(x) = 0 whenever x > 2.

6. It should be rather clear from the plot that

\displaystyle \lim_{x\to0}f(x) = \lim_{x\to0}(\sin(x)+3) = \sin(0) + 3 = \boxed{3}

because sin(x) + 3 is continuous at x = 0. On the other hand, the limit at infinity doesn't exist because sin(x) oscillates between -1 and 1 forever, never landing on a single finite value.

For 7-8, divide through each term by the largest power of x in the expression:

7. Divide through by x². Every remaining rational term will converge to 0.

\displaystyle \lim_{x\to\infty}\frac{x^2+x-12}{2x^2-5x-3} = \lim_{x\to\infty}\frac{1+\frac1x-\frac{12}{x^2}}{2-\frac5x-\frac3{x^2}}=\boxed{\frac12}

8. Divide through by x² again:

\displaystyle \lim_{x\to-\infty}\frac{x+3}{x^2+x-12} = \lim_{x\to-\infty}\frac{\frac1x+\frac3{x^2}}{1+\frac1x-\frac{12}{x^2}} = \frac01 = \boxed{0}

9. Factorize the numerator and denominator. Then bearing in mind that "x is approaching 6" means x ≠ 6, we can cancel a factor of x - 6:

\displaystyle \lim_{x\to6}\frac{2x^2-12x}{x^2-4x-12}=\lim_{x\to6}\frac{2x(x-6)}{(x+2)(x-6)} = \lim_{x\to6}\frac{2x}{x+2} = \frac{2\times6}{6+2}=\boxed{\frac32}

10. Factorize the numerator and simplify:

\dfrac{-2x^2+2}{x+1} = -2 \times \dfrac{x^2-1}{x+1} = -2 \times \dfrac{(x+1)(x-1)}{x+1} = -2(x-1) = -2x+2

where the last equality holds because x is approaching +∞, so we can assume x ≠ -1. Then the limit is

\displaystyle \lim_{x\to\infty} \frac{-2x^2+2}{x+1} = \lim_{x\to\infty} (-2x+2) = \boxed{-\infty}

6 0
2 years ago
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