A , O and B lie on a straight line segment

Evaluate x the diagram is not drawn to scale

This question is about exponential growth and the rabbit population on an island.

The subject of this question is Mathematics. Specifically, it involves exponential growth and a function that represents the rabbit population on the island over time. Let's define the function f(x) to represent the population of rabbits at year x.

Based on the information given, the initial population is 20 rabbits. The population triples every year, which means that for each year x, the population is f(x) = 20 * 3^x.

For example, after 1 year, the population would be f(1) = 20 * 3^1 = 60 rabbits. After 2 years, the population would be f(2) = 20 * 3^2 = 180 rabbits, and so on.

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Answer:

[tex]\text{Solution: }\left(-\dfrac{2654}{435},\dfrac{1644}{145}\right)[/tex]

Step-by-step explanation:

Line P is drawn by joining ordered pairs (-8,15) and (6,-12)

Using two point formula of line. Find equation of line P

[tex]\dfrac{y-15}{x-(-8)}=\dfrac{-12-15}{6-(-8)}[/tex]

[tex]\dfrac{y-15}{x+8}=\dfrac{-27}{6+8)}[/tex]

[tex]\dfrac{y-15}{x+8}=\dfrac{-27}{14}[/tex]

[tex]y=\dfrac{-27}{14}(x+8)+15[/tex]

[tex]y=\dfrac{-27}{14}x+\dfrac{-27}{14}\times 8 + 15[/tex]

[tex]y=\dfrac{-27}{14}x-\dfrac{3}{7}[/tex]

[tex]\text{Equation of line P:}y=\dfrac{-27}{14}x-\dfrac{3}{7}[/tex]

Line Q is drawn by joining ordered pairs (4,16) and (-9,10)

Using two point formula of line. Find equation of line Q

[tex]\dfrac{y-16}{x-4}=\dfrac{10-16}{-9-4}[/tex]

[tex]\dfrac{y-16}{x-4}=\dfrac{-6}{-13}[/tex]

[tex]\dfrac{y-16}{x-4}=\dfrac{6}{13}[/tex]

[tex]y=\dfrac{6}{13}(x-4)+16[/tex]

[tex]y=\dfrac{6}{13}x-\dfrac{6}{13}\times 4 + 16[/tex]

[tex]y=\dfrac{6}{13}x+\dfrac{184}{13}[/tex]

[tex]\text{Equation of line Q:}y=\dfrac{6}{13}x+\dfrac{184}{13}[/tex]

Using graphing find solution of system of equation.

We draw the line on graph and see the intersection point.

Please see the attachment of the line.

Answer:

(−2, 4), because it is the point of intersection of the two graphs

Step-by-step explanation:

BECAUSE

Kevin and Randy have 50 quarters and 18 nickels.

To solve this problem, we can set up a system of equations. Let's use the variables x for the number of quarters and y for the number of nickels. We can then write two equations based on the given information:

Equation 1: x + y = 68 (the total number of coins is 68)

Equation 2: 0.25x + 0.05y = 13.40 (the total value of the coins is $13.40).

We can start by solving Equation 1 for x in terms of y:

x = 68 - y. Now we can substitute this expression for x into Equation 2:

0.25(68 - y) + 0.05y = 13.40

Simplifying, we get 17 - 0.25y + 0.05y = 13.40.

Combining like terms, we have 0.20y = 3.60.

Dividing both sides by 0.20, we find y = 18. Plugging this value back into Equation 1, we can solve for x:

x + 18 = 68

x = 68 - 18 = 50.

Therefore, Kevin and Randy have 50 quarters and 18 nickels.

A random sample is a subset of a population where every individual has an equal and known chance of being selected, providing a representative segment for study.

A random sample is a sample in which each member of the population has a known, nonzero, chance of being selected for the sample. This means that in a random sample, every individual in the larger population has an equal opportunity to be included. There are various methods to achieve a random sample, such as simple random sampling, stratified sampling, or using random digit dialing. Random samples are crucial because they tend to be representative of the population, thereby allowing the results of studies to be generalizable.

Correct Answer is B which is $ 358.47 for monthly payment

The population will be approximately 4,437 fish after 5 years.

To predict the fish population 5 years after the lake is stocked,

We can use the formula for exponential decay:

[tex]\[ P = P_0 \times (1 - r)^t \][/tex]

Where:

Given:

Substituting the values into the formula:

[tex]\[ P = 10,000 \times (1 - 0.15)^5 \][/tex]

[tex]\[ P = 10,000 \times (0.85)^5 \][/tex]

Calculating:

P ≈ 10,000 × 0.443705

P ≈ 4437.05

So, the fish population 5 years after the lake is stocked is approximately 4437 fish.

To calculate the total amount and interest earned on $6,500 at 6% simple interest for 25 years, you use the formula I = P × r × t, which gives you interest (I) of $9,750. Adding the interest to the principal, the total amount after 25 years is $16,250.

To calculate the total amount and the interest earned on $6,500 at a simple interest rate of 6% for 25 years, we can use the simple interest formula:

I = P × r × t

Where:

Using this formula:

I = $6,500 × 0.06 × 25

I = $9,750

The total amount after 25 years will be the principal plus the interest:

Total Amount = P + I

Total Amount = $6,500 + $9,750

Total Amount = $16,250

The interest earned on the investment after 25 years is $9,750.

The correct option is (C)

Here, we need to find the length of the hose which is diagonally placed in the rectangular garden.

The length is given as = 72 meters

The width is given as = 60 meters

We will use Pythagoras theorem here.

Hypotenuse² = Perpendicular² + Base²

Hypotenuse² = 72² + 60²

Hypotenuse² = √8,784 meters²

Hypotenuse = 93.72 meters

Thus, the length of diagonal hose will be = 93.72 meters

Answer:

Option B. 49.5 degrees

Step-by-step explanation:

Given that arc QS measures 4x-18

Arc QS is a semicircle and hence subtended angle by semicircle at the centre is 180 degrees.

Use this to find the value of x

4x-18 = 180 degrees

Add 18 to both sides

4x=198

Divide by 4

x =198/4 =49.5

So answer is option B

49.5 degrees

Answer:

49.5

Step-by-step explanation:

Answer:

B) -1

Step-by-step explanation:

The ball hits the ground after approximately 4.16 seconds.

To find the time at which the ball strikes the ground, we need to determine when h(t) = 0. Given the equation for height:

[tex]\[ h(t) = -16t^2 - 21t + 190 \][/tex]

we can set h(t) to 0 and solve for t:

[tex]\[ -16t^2 - 21t + 190 = 0 \][/tex]

Using the quadratic formula, we can find the roots of this equation:

[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

where a = -16 , b = -21, and c = 190. Plugging in the values, we get:

1. Find the discriminating value [tex]\( \Delta = b^2 - 4 \times a \times c \)[/tex]:

[tex]\[ \Delta = (-21)^2 - 4 \times (-16) \times 190 = 441 + 12160 = 12601 \][/tex]

2. Calculate the roots:

[tex]\[ t = \frac{-(-21) \pm \sqrt{12601}}{2 \times -16} \\ \[ t = \frac{21 \pm \sqrt{12601}}{-32} \][/tex]

Since we're interested in the positive value (representing the time when the ball hits the ground), we'll only consider the root with the "minus" sign:

[tex]\[ t = \frac{21 - \sqrt{12601}}{-32} \approx \frac{-21 + \sqrt{12601}}{32} \approx \frac{21 - \sqrt{12601}}{32} \][/tex]

Using a calculator, let's approximate the result:

[tex]1. Compute \( \sqrt{12601} \approx 112.24 \),\\2. Then calculate \( 21 + 112.24 \approx 133.24 \),\\3. Divide by 32 to get \( t \approx \frac{133.24}{32} \approx 4.16375 \).[/tex]

Rounding to a more readable result, we can say that the ball hits the ground after approximately 4.16 seconds.