Estimate the average annual growth rate of the world population for 1950, 1988, and 2010 using model 17 from the previous page.Model 17. Year Population Growth Rate (%)
1972 1.96
1982 1.73
1992 1.5
2002 1.22

Answers

Answer 1

Answer: although I don't see 1950, 1988, or 2010 on your model, I did calculate the average for 1972, 1982, 1992, and 2002: the average that I got was 1.6025 (this answer has not been rounded due to the fact that I don't know if you need to round it)
hope this helps

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A polynomial function has a root of –6 with multiplicity 1, a root of –2 with multiplicity 3, a root of 0 with multiplicity 2, and a root of 4 with multiplicity 3. If the function has a positive leading coefficient and is of odd degree, which statement about the graph is true?A)The graph of the function is positive on (–6, –2). B)The graph of the function is negative on (negative infinity, 0). C)The graph of the function is positive on (–2, 4). D)The graph of the function is negative on (4, positive infinity).
Complete the inequality statement. (help)5/11___7/11
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PLEASE SOLVE

Solve for x if x+9=7

Answers

Answer:

-2 is the correct answer

Answer:

x = -2

Step-by-step explanation:

x + 9 = 7

=> x = 7 - 9

=> x = -2

Gary bought a car for $40,000. If V = 40,000(.85)t represents the value of the car after t years, how long will it take the car to be worth less than one-fourth of its purchase price? A) 4 years B) 6 years C) 8 years D) 9 years

Answers

Answer:

D) 9 years.

Step-by-step explanation:

We have been given that Gary bought a car for $40,000 and equation $V=40,000(0.85)^t$ represents the value of the car after t years.

First of all we will find the one-fourth of 40,000.

$\text{One-forth of car's purchase price}=(\$40,000)/(4)$

$\text{One-forth of car's purchase price}=\$10,000$

To find the time it will take the car to be worth less than one-fourth of its purchase price, we will substitute V=10,000 in our given equation.

$10,000=40,000(0.85)^t$

Let us divide both sides of our equation by 40,000.

$(10,000)/(40,000)=(40,000(0.85)^t)/(40,000)$

$0.25=0.85^t$

Let us take natural log of both sides of our equation.

$ln(0.25)=ln(0.85^t)$

Using natural log property $ln(a^b)=b*ln(a)$ we will get,

$ln(0.25)=t*ln(0.85)$

$(ln(0.25))/(ln(0.85))=(t*ln(0.85))/(ln(0.85))$

$(-1.3862943611198906)/(-0.1625189294977749)=t$

$8.530048563597=t$

Upon rounding our answer to the nearest year we will get,

$t\approx 9$

Therefore, it will take 9 years the car to be worth less than one-fourth of its purchase price and option D is the correct choice.

What is the next in the sequence 2 6 3 8 6 12

Answers

The correct answer is: 11

Explanation:
There are two series in the given sequence:

Series-1:
2 3 6 x

Where x represents the number we need to find.

Series-2:
6 8 12

The following pattern is seen in the Series-2:
6 + 2 = 8
8 + 4 = 12

Meaning the additive has the incremental increase of 2 [Like 2, 4, 6, 8 ...].

Likewise,
In Series-1, the following pattern is shown [1,3,5,7...]:
2 + 1 = 3
3 + 3 = 6
6 + 5 = 11

Hence the correct answer is 11.

11
As you can see, the pattern is subsequently +4, -3, +5, -2, +6, -1 etc.
So, according to this, next sequence should be 12-1=11

-x-y=11
convert from standard form to slope intercept form

Answers

Answer:

y =-x-11

Step-by-step explanation:

Slope intercept form : y=mx+b

$-x-y=11\n\n-y =x +11$

Multiply by -1(reverse the inequality)

$-1(-y=x+11)\n\ny =-x-11$

Answer:

y = - x - 11

Step-by-step explanation:

-x-y=11

+x +x

-y=11+x

/- /-

y=-11-x

rearrange

y = - x - 11

PLEASE HELP! 80 POINTS AND BRAINLIEST TO FIRST ANSWER! A rectangular pen is constructed with 300m of fencing against a barn wall. What dimensions give the maximum area?

Answers

I want to say a 100m on each side.

since it’s against a barn wall, then there would be three sides left. But I could be wrong. ¯\_(ツ)_/¯

Discriminant to determine the nature of the roots of x2 + 2x + 5 = 0. a) No real roots
b) One real root
c) Two distinct real roots

Answers

discrimant = b^2 - 4ac = 2^2 - (4 x 1 x 5) = 4 - 20 = -16
Therefore, there is no real roots.

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Estimate the average annual growth rate of the world population for 1950, 1988, and 2010 using model 17 from the previous page.Model 17. Year Population Growth Rate (%)1972 1.961982 1.731992 1.52002 1.22

Answers

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y =-x-11

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Estimate the average annual growth rate of the world population for 1950, 1988, and 2010 using model 17 from the previous page.Model 17. Year Population Growth Rate (%)
1972 1.96
1982 1.73
1992 1.5
2002 1.22