Bruce bought a map that cost $7.35 he used a $10 bill to pay for his map what is the change

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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Answer: quadrant 4

Step-by-step explanation:

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.

Answer:

{(3, –1), (7, 1), (–6, –1), (9, 1), (2, –1)}

Step-by-step explanation:

A set of ordered pairs in the format [tex](x,y)[/tex] represents a function if for each value of x, there is only one value for y.

The first set of ordered pairs is:

{(2, –2), (1, 5), (–2, 2), (1, –3), (8, –1)}

There are two values of y for [tex]x = 1[/tex]. This means that this set of ordered pairs does not represent a function.

The second set of ordered pairs is:

{(3, –1), (7, 1), (–6, –1), (9, 1), (2, –1)}

For each value of x here, there is only one value of y. This means that this set of ordered pairs represents a function. This is the answer.

The third set of ordered pairs is:

{(6, 8), (5, 2), (–2, –5), (1, –3), (–2, 9)}

There are two values of y for [tex]x = -2[/tex]. This means that this set of ordered pairs does not represent a function.

The fourth set of ordered pairs is:

{(–3, 1), (6, 3), (–3, 2), (–3, –3), (1, –1)}

There are two values of y for [tex]x = -3[/tex]. This means that this set of ordered pairs does not represent a function.

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

To find the radius of a circle with an area of 254 cm², you can use the formula for the area of a circle by solving for the radius. We get the radius as 8.99 cm.

The radius of a circle can be found by using the formula for the area of a circle:

Here, we need to find unit cost. Unit cost can be defined as a cost per unit of a product.

We are given that 16 ounce will cost $ 4.00.

That means -

Cost of 16 ounce of cereal package = $ 4.00

Cost of 1 ounce of cereal package -

(We will divide both sides by 16)

Cost of 16 ounce of cereal package ÷ 16 = $ 4.00 ÷ 16

Cost of 1 ounce of cereal package = $ 0.25.

Thus, the cost of 1 ounce of cereal package = $ 0.25.

Answer: 1 oz. Over 0.25 is your answer.

Step-by-step explanation:

Final answer:

For the expression R(x)=(500+ax)(50-bx), a represents the increase in yearbooks sold for each $5 decrease in price and b represents the corresponding decrease in price per yearbook. The correct values for a and b are 100 and 5 respectively, resulting in the revenue expression R(x) = (500 + 100x)(50 - 5x).

Explanation:

The student's question revolves around the concept of revenue generation based on the number of items sold and the selling price per item.

When examining revenue, we understand it as the product of price per unit times the number of units sold. In the given example, a school is selling yearbooks with a base scenario of 500 yearbooks at $50 each. We are given that for every $5 decrease in price, the school can sell an additional 100 yearbooks. This can be formulated as:

R(x) = (500 + ax)(50 - bx)

Here x represents the number of $5 decreases from the original price. With each decrease, the number of books sold increases by 100, and the price decreases by $5.

Therefore, variable a must represent the increase in quantity sold for each $5 price decrease (a = 100), and variable b must represent the decrease in price per yearbook for each price decrease (b = 5).

Therefore, the values of a and b are 100 and 5, respectively.

The expression for the revenue becomes:

R(x) = (500 + 100x)(50 - 5x)

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:

Answer:

d

Step-by-step explanation:

denisity * volume = mass

Answer: I, III, and IV only

Step-by-step explanation:

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