jenny is diving into a swimming pool from a diving board. The expression below represents Jenny's vertical height, in feet, above the surface of the water after x seconds. -16x^2 +12x+10. Which statement best describes the term 10?

A. Jenny's initial height above the surface of the water
B. The time it takes Jenny to reach the surface of the water
C. The horizontal distance from the diving board Jenny travels
D. the maximum height Jenny reaches.

Answer: (40+10π) mm.

Step-by-step explanation: Given diameter of the circle = 10mm.

Radius of the circle = 10/2 = 5 mm.

From the figure, we can see that each or the four corners makes a quarter of the circle.

And four corners would make the complete circle with radius 5mm.

Therefore, circumference of the four quarters of a circle = 2 π r

= 2 π  (5) = 10 π.

Now, we can see that each side 10 mm of band also there.

So, the length of straight band on all four sides = 4 × 10 = 40 mm.

Total length of the band = (40+10π) mm

The length of the band is the number of units it covers

The actual length of the band is: [tex]\mathbf{40 + 10\pi}[/tex]

The diameter of 1 pencil is given as:

[tex]\mathbf{d = 10}[/tex]

The circumference of 1 pencil is:

[tex]\mathbf{C = \pi d}[/tex]

So, we have:

[tex]\mathbf{C = 10\pi }[/tex]

Multiply by 4, to calculate the circumference of all pencils

[tex]\mathbf{4C =4 \times 10\pi }[/tex]

[tex]\mathbf{4C =40\pi }[/tex]

The rubber band covers a quarter of the 4 pencils.

So, we divide by 4

[tex]\mathbf{C = 10\pi }[/tex]

The length of the band on one pencil is:

[tex]\mathbf{L = 10}[/tex]

Multiply by 4, for the 4 pencils

[tex]\mathbf{L = 40}[/tex]

So, the actual length of the band is:

[tex]\mathbf{Length = 40 + 10\pi}[/tex]

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The above answer will be calculated as -

We are given -

600 ( 1 + 0.03)^12

We will first add - 1 + 0.03 = 1.03

Now, we will multiply 1.03 into 12 times, in order to find the value of ( 1.03)^12

The answer for ( 1.03)^12 = 1.4268

Now, we will multiply 1.4268 with 600

600 × 1.4268 = 856.08 approximately.

Thus, the value of 600*(1.03)^12 has been calculated.

Answer: I, III, and IV only

Step-by-step explanation:

The standard form of the given number is 3.89936206767 × 10¹¹

From the question, the given number is

three hundred eighty-nine billion, nine hundred thirty-six million, two hundred six thousand, seven hundred sixty-seven.

First, we will write the number in figures before writing it in standard form.

The number, three hundred eighty-nine billion, nine hundred thirty-six million, two hundred six thousand, seven hundred sixty-seven, written in figure is

389,936,206,767.

Now, to write in standard form,

A number is said to be in standard form, if it is written in form A × 10ⁿ.  Where A is a number bigger than or equal to 1 and less than 10 ( That is, A ≥ 1 and A < 10)

and n is any positive or negative whole number

∴ The number, 389,936,206,767, written in standard form is 3.89936206767 × 10¹¹

Hence, the standard form of the given number is 3.89936206767 × 10¹¹

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The cost of the automobile, excluding the impact of the interest, is $26,500.

For this question, we need to determine the total cost of the automobile purchased by Dave Krug. We know that Dave made an upfront payment of $6,500, and also agreed to make 40 monthly payments of $500 each. The total amount paid through monthly payments is $500 * 40 = $20,000.

Here, it is also stated that the loan bears an annual interest rate of 12%. However, no additional data are given for us to calculate the impact of the interest on the overall cost of the car. So, we'll proceed without including the interest.

So, to find the total cost of the car, we combine the upfront payment with the total monthly payments. That is, $6,500 (upfront payment) + $20,000 (total monthly payments) = $26,500. Therefore, the cost of the automobile without taking into account the interest is $26,500.

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For every mole of carbon dioxide used, one mole of methane is produced. Therefore, 71.1 moles of carbon dioxide would produce 71.1 moles of methane, assuming hydrogen is in excess.

According to the given reaction, we can infer that the stoichiometric ratio between methane (CH4) and carbon dioxide (CO2) is 1:1. This means for each mole of methane produced, one mole of carbon dioxide is consumed. Therefore, if you are starting with 71.1 moles of carbon dioxide and assuming there is excess hydrogen, you would produce 71.1 moles of methane because of this 1:1 mole ratio.

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The angles for Brian's triangular plot of farmland are approximately 27° for ∠A, 38° for ∠B, and 115° for ∠C.

To determine the angles of a triangular plot of land with sides measuring 1.1 miles, 1.5 miles, and 2.2 miles, we can use the Law of Cosines. The Law of Cosines states:

⇒ c² = a² + b² - 2ab × cos(C)

⇒ a = 1.1 miles

⇒ b = 1.5 miles

⇒ c = 2.2 miles

⇒ cos(C) = (a² + b² - c²) ÷ (2ab)

⇒ cos(C) = (1.1² + 1.5² - 2.2²) ÷ (2 × 1.1 × 1.5)

= (1.21 + 2.25 - 4.84) ÷ (3.3)

= (-1.38) ÷ (3.3)

= -0.4182

⇒ ∠C ≈ 115°

⇒ cos(A) = (b² + c² - a²) ÷ (2bc)

⇒ cos(A) = (1.52 + 2.22 - 1.12) ÷ (2 × 1.5 × 2.2)

= (2.25 + 4.84 - 1.21) ÷ (6.6)

= (5.88) ÷ (6.6)

= 0.8909

⇒ ∠A ≈ 27°

⇒ ∠B = 180° - ∠A - ∠C

= 180° - 27° - 115°

= 38°

The angles at which the fences of the three sides will meet are approximately 27° for ∠A, 38° for ∠B, and 115° for ∠C, when rounded to the nearest degree.

Complete question:

Brian wants to fence in his triangular plot of farm land that measures 1.1 × 1.5 × 2.2 miles.

Determine the angles at which the fences of the three sides will meet.

Rounding each angle to the nearest degree:

Here, in order to find the solutions to the quadratic equation - x² + x - 30 = 0, we will use factorization method.

In the method, we will split 30, in such factors, which when added or subtracted gives us 1,  and when multiplied gives us -30.

So, we will use, -5 and 6. When they are added they will give us 1 and when multiplied they will give us -30 as answer.

Now, the equation will be written as -

x² - 5x + 6x - 30 = 0

Taking common, we get

x(x - 5) +6(x-5) = 0

(x-5)(x+6) = 0

So, x - 5 = 0 and x +6 = 0, we will get, x = 5 and x = - 6

Thus, the correct option is C). x = 5 and -6

The simplification of ( n + 8) + ( n -12 ) is 2(n -2)

algebraic expression are expression with unknown.

This unknown is represented by letters and in linear algebraic equations , the highest power of the unknown is 1.

Simplifying (n + 8) + (n-12)

n + 8 + n -12

collecting like terms

n + n + 8 -12

2n -4

we can further Factorize

2( n -2)

Therefore, the simplification of ( n + 8) + ( n -12 ) is 2(n -2)

the correct option is:

b.

Step 1:0.035 xx16=0.56

Step 2: 16-0.56=15.44

Step 3: 0.0425 xx15.44=0.6562

Step 4: 15.44-0.6562=14.7838

Let's break down the calculations:

Step 1: Calculate the amount of water lost at the end of the first hour.

[tex]\[ \text{Amount lost} = 0.035 \times 16 = 0.56 \][/tex]

Step 2: Subtract the amount lost from the initial amount to find the remaining water at the end of the first hour.

[tex]\[ \text{Remaining water} = 16 - 0.56 = 15.44 \][/tex]

Step 3: Calculate the amount of water lost at the end of the second hour.

[tex]\[ \text{Amount lost} = 0.0425 \times 15.44 = 0.6562 \][/tex]

Step 4: Subtract the amount lost from the remaining water at the end of the first hour to find the remaining water at the end of the second hour.

[tex]\[ \text{Remaining water} = 15.44 - 0.6562 = 14.7838 \][/tex]

So, the correct option is:

b.

Step 1:0.035 xx16=0.56

Step 2: 16-0.56=15.44

Step 3: 0.0425 xx15.44=0.6562

Step 4: 15.44-0.6562=14.7838

The complete Question is given below:

A student uses a solution that contains 16 grams of water to conduct an evaporation experiment.

At the end of one hour, the amount of water in the solution has decreased by 3.5% .

At the end of two hours, the amount of water in the solution has decreased by another 4.25% .

Which calculations can be used to determine the amount of water, in grams, remaining in the solution at the end of the second hour?

a.

Step 1:0.035 xx16=0.56

Step 2: 16-0.56=15.44

Step 3: 0.0425 xx15.44=0.6562

Step 4:16-0.6562=15.3438

b.

Step 1:0.035 xx16=0.56

Step 2: 16-0.56=15.44

Step 3: 0.0425 xx15.44=0.6562

Step 4: 15.44-0.6562=14.7838

c.

Step 1:0.35 xx16=5.6

Step 2: 16-5.6=10.4

Step 3: 0.425 xx10.4=4.42

Step 4:16-4.42=11.58

d.

Step 1: 0.35 xx16=5.6

Step 2: 16-5.6=10.4

Step 3: 0.425 xx10.4=4.42

Step 4:10.4-4.42=5.98

The equation relating females population with time is [tex]\boxed{t =  - \ln \left| P \right| + 6\ln \left| {0.2P - 400} \right| - 12\ln \left( 5 \right) + 4\ln \left( {10} \right) - 36\ln \left( 2 \right)}.[/tex]

Further explanation:

Explanation:

The female population is related to time and it is given as follows,

[tex]t = \int {\dfrac{{p + s}}{{p\left[ {\left( {r - 1} \right)p - s} \right]}}dp}[/tex]

The sterile males are 400 and the growth rate is 1.2.

[tex]\begin{aligned}t&= \int {\frac{{p + 400}}{{p\left[ {\left( {1.2 - 1} \right)p - 400} \right]}}dp}\\&= \int {\frac{{p + 400}}{{p\left[ {0.2p - 400} \right]}}dp}\\\end{aligned}[/tex]

The expression [tex]\dfrac{{p + 400}}{{p\left[ {0.2p - 400} \right]}}[/tex] can also be expressed as follows,

[tex]\dfrac{{p + 400}}{{p\left[ {0.2p - 400} \right]}} = \dfrac{A}{p} + \dfrac{B}{{\left( {0.2p - 400} \right)}}[/tex]

Find the least common denominator.

[tex]\dfrac{{p + 400}}{{p\left( {0.2p - 400} \right)}} = \dfrac{{A\left( {0.2p - 400} \right)}}{{p\left( {0.2p - 400} \right)}} + \dfrac{{Bp}}{{p\left( {0.2p - 400} \right)}}[/tex]

[tex]\begin{aligned}p + 400 &= 0.2Ap - 400A + Bp\\p + 400 &= p\left( {0.2A + B} \right) - 400A\\\end{aligned}[/tex]

Equate left and right side of the expression to obtain the value of A and B.

[tex]\begin{aligned}0.2A + B &= 1\\- 400A &= 400\\\end{aligned}[/tex]

The value of A can be obtained as follows,

[tex]\begin{aligned}A&= \dfrac{{400}}{{ - 400}}\\A& =- 1\\\end{aligned}[/tex]

The value of B is 1.2.

Integral can be expressed as follows,

[tex]\begin{aligned}t &= \int {\frac{{p + 400}}{{p\left( {0.2p - 400} \right)}}dp}  \\ &= \int {\left( {\frac{{ - 1}}{p} + \frac{{1.2}}{{\left( {0.2p - 400} \right)}}} \right)dp} \\&= - \int {\frac{1}{p}dp}  + 6 \times \int {\frac{{0.2}}{{0.2p - 400}}dp}\\&= - \ln \left| p \right| + 6\ln \left| {0.2p - 400} \right| + C\\\end{aligned}[/tex]

At initial time the population is 10000.

[tex]\begin{aligned}- \ln \left| {10000} \right| + 6\ln \left| {0.2 \times 10000 - 400} \right| + C &= 0\\\ln \left| {10000} \right| - 6\ln \left| {1600} \right| &= C\\4\ln 10 - 36\ln \left( 2 \right) - 12\ln \left( 5 \right) &= C\\\end{aligned}[/tex]

Substitute [tex]4\ln 10 - 36\ln \left( 2 \right) - 12\ln \left( 5 \right)[/tex] for [tex]C.[/tex]

[tex]t =  - \ln \left| p \right| + 6\ln \left| {0.2p - 400} \right| + 4\ln 10 - 36\ln \left( 2 \right) - 12\ln \left( 5 \right)[/tex]

The equation relating females population with time is [tex]\boxed{t = - \ln \left| P \right| + 6\ln \left| {0.2P - 400} \right| - 12\ln \left( 5 \right) + 4\ln \left( {10} \right) - 36\ln \left( 2 \right)}.[/tex]

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Answer details:

Grade: High School

Subject: Mathematics

Chapter: Exponential function

Keywords: exponential growth, growth rate model, pesticides, slowing, insect population, sterile males, mate, fertile, females, explicitly, female grows, integral, female insects, male insects, capita rate, chosen mate, generation.

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