1/8 of the toys in a play area are yellow, and 1/4 are green. 3/5 of the remainder are red. The rest of the toys are blue. If there are 100 green toys, how many blue toys are there?​

The table shows the amounts of each day is attached below.

An equation is an expression that shows the relationship between two or more numbers and variables.

The total amount of snow that falls in three days equals the sum of the amount of snow that falls each day, then

s1 + s2 + s3 = Ts

Here s1 is the snow that fell on day 1, s2 on day 2, s3 on day 3 and Ts is the total amount of snow that fell over the three days.

We need to calculate the amount of snow that fell on the first two days by multiplying the total snow by the given fractions, like this:

s1 = 1/6 * 2 = 1/3

s2 = 7/12*2 = 7/6

Now ,

1/3+ 7/6+ s3 = 2

1.5 + s3 = 2

1.5 - 1.5 + s3 = 2 - 1.5

s3 = 2 - 1.5

s3 = 1/2

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Answer:

Let's call the number of magazines charged at £1.90 each "x".

So, the number of magazines charged at £1.70 each is "t=70-x".

The total cost of the x magazines is x * £1.90, and the total cost of the t magazines is t * £1.70.

The total bill for all 70 magazines is £124, so we can write an equation:

x * £1.90 + t * £1.70 = £124

x * £1.90 + (70-x) * £1.70 = £124

Expanding the equation, we get:

x * £1.90 + 70 * £1.70 - x * £1.70 = £124

x * £1.90 - x * £1.70 = £124 - 70 * £1.70

x * (£1.90 - £1.70) = £124 - 70 * £1.70

x * £0.20 = £124 - 70 * £1.70

x = (£124 - 70 * £1.70) / £0.20

Now that we have an expression for x, we can use a calculator to solve for it.

x = (124 - (70 * 1.70)) / 0.20 = 40.

So, 40 magazines were charged at £1.90 each and 70 - 40 = 30 magazines were charged at £1.70 each.

The perimeter of the second park will be 320 feet. The correct option is C.

The perimeter is defined as the sum of all the sides of the object. For a rectangular park, the perimeter will be the sum of all the sides of the rectangular park.

Given that a  park has two rectangular, fenced playgrounds. The first playground has a perimeter of 160 feet.

The perimeter of the first park:-

160 = 2 ( L + W )

L + W = 80

The perimeter of the second park,

P = 2 ( 2L + 2W )

P = 4 ( L + W )

P = 4 x 80

P = 320 feet

Therefore, the perimeter will be 320 feet.

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Answer:

reflection  over the x axis

vertical stretch of 2

vertical translation down 1 unit

Step-by-step explanation:

-2 [tex]\sqrt[3]{x}[/tex] -1

The negative in front is a reflection  over the x axis

The 2  is a vertical stretch of 2

The subtraction of 1 at the end is  a vertical translation down 1 unit

Answer:

k = 5.6

Step-by-step explanation:

[tex]\frac{10}{7} = \frac{8}{k}[/tex]

Cross-multiplication is applied:

(10)(k) = (8)(7)

   10k = 56

k needs to be isolated and made the subject of the formula:

k = [tex]\frac{56}{10}[/tex]

k = 5.6

Answer: Alicia had 3 feet of wood and she used 3 feet, so she has 3 - 3 = 0 feet of wood left.

Since we are asked to estimate the amount of wood, we could round 3 to the nearest whole number and say that Alicia has approximately 0 feet of wood left.

Step-by-step explanation:

Answer: The equation that represents the relationship between the price of oranges last week, x, and the price of oranges this week, y is:

y = x + 1/5x

The equation states that the price of oranges this week is equal to the price of oranges last week plus 1/5 of the price of oranges last week.

Step-by-step explanation:

The definite integral of a continuous function on an interval [a, b] represents the area under the graph of the function within that interval. It provides a mathematical way to measure the accumulation of change of the function over that interval.

The definite integral of a function f on the interval [a, b] is denoted by ∫f(x)dx over the interval [a, b]. This integral can be approximated by dividing the interval [a, b] into smaller subintervals and calculating the area under the graph of the function over each subinterval. The smaller the subintervals, the closer the approximation will be to the actual definite integral.

The definite integral is said to be a limit, meaning that as the size of the subintervals approaches zero, the sum of the areas under the graph over each subinterval approaches a definite value, which is the definite integral.

In mathematical terms, for each positive number e, there exists a positive number δ such that if the subintervals have width less than δ, then the difference between the sum of the areas under the graph and the definite integral will be less than e. This means that the definite integral provides a precise value for the area under the graph of the function over the interval [a, b].

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In the image you can see some rectangles with different dimensions, to scale.

A rectangle is a geometric figure that is characterized by:

To find the perimeter of a rectangle we must add all its sides. In this case we must illustrate a 20 m rectangle so we can draw the following variants:

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The most precise classification of BCDE is a square.

Quadrilaterals are four sided polygons which also have four vertices and four angles.

Sum of all the interior angles of a quadrilateral is 360 degrees.

We have to find the length of adjacent sides of the quadrilateral BCDE.

If they are equal, then the quadrilateral is either square or rhombus.

If they are not equal, then they are either rectangle or parallelogram.

B(-1, -1), C(6, -2), D(5, -9), and E(-2, -8)

BC = [tex]\sqrt{(6--1)^2+(-2--1)^2}[/tex] = √50

CD = [tex]\sqrt{(5-6)^2+(-9--2)^2}[/tex] = √50

Adjacent sides are equal. So it is either square or rhombus.

If it is rhombus, length of diagonals BD and CE are not equal.

If it is square, length of diagonals BD and CE are equal.

BD = [tex]\sqrt{(5--1)^2+(-9--1)^2}[/tex] = √100 = 10

CE = [tex]\sqrt{(-2-6)^2+(-8--2)^2}[/tex] = √100 = 10

Length of diagonals are same. So it is a square.

Hence the given quadrilateral is a square.

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Answer:

The $19,000 in expenses should be entered by the accountant as a debit in the following journal entry:

Date Account Account Number Debit Credit

1 - Oct Expenses 19000 5050

The value of the given expression is -6

What is an expression?

Mathematical expressions consist of at least two numbers or variables, at least one arithmetic operation, and a complete sentence. Any one of the following mathematical operations can be used. A sentence has the following structure: Number/variable, Math Operator, Number/Variable is an expression.

f(-2) value is the value of f(x) when x= -2

From the graph ,

f(-2)= 0

g(7)=2

Given:

 -5 * f(-2) - 3*g(7)

= -5*0 - 3*2

First multiplication ,then subtraction from left to right

= 0- 3*2

=0-6

= -6

So, the value of the given expression is -6

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Using the midpoint formula we will see that the coordinates of B are (-9, 1)

We know that for two points (x, y) and (a, b) the midpoint is at:

( (x  + a)/2,  (y + b)/2)

In this case if the coordinates of B are (x, y), we can write the equation:

(-2, 4) = ( (x + 5)/2, (7 + y)/2)

We can solve these two equations to find the coordinates of point B, we will get:

(x + 5)/2 = -2

x + 5 = -4

x = -4 - 5 = -9

And:

(7 + y)/2 = 4

7 + y = 4*2 = 8

y = 8 - 7 = 1

The coordinates of B are (-9, 1)

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The presented statement states that the economic profit at units has a numerical value of -512 dollars per unit.

A function is a collection of several statements that work together to complete a job. Every C program we use contains at least one function, which is main (). The programs can then declare a number of extra routines in a language.

Revenue function is R (q) = 7q² + 400q

Differentiate with respect to q

R'(q) = - 7(2q) + 400.......(1)

R'(q) = - 14q + 400

Substitute q = 8 in eqn.(1)

MR(8) = -14(8) + 400

= - 112 + 400

MR(8) = -512 $ per unit

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The required explanation for hypothesis test is described.

When the null hypothesis is true in this setting, it means that the true proportion of students at Mr. Tabor's school who completed their math homework last night is equal to 0.75, or 75%. This is the value that Mr. Tabor initially believed, but the sample data suggests that only 68% of the students completed their homework.

A PP-value is the probability of observing a sample proportion as extreme or more extreme than the sample proportion, given that the null hypothesis is true. In this case, the PP-value of 0.1265 is not small enough to reject the null hypothesis. A commonly used significance level is 0.05, which means that if the PP-value is less than 0.05, the null hypothesis would be rejected.

So, given the PP-value of 0.1265, which is greater than 0.05, we cannot reject the null hypothesis and must conclude that there is not enough evidence to suggest that the true proportion of students who completed their homework is less than 0.75. This means that, based on the sample data, the null hypothesis that the true proportion is equal to 0.75 is not ruled out, and there is not enough evidence to support the alternative hypothesis that the true proportion is less than 0.75.

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The second largest of the four angles is angle ∠1

Triangle is a closed polygon that is formed by connecting 3 line segments which are called sides of the triangle. In a triangle, the shortest side will always be placed opposite the smallest interior angle. Similarly, the longest side will always be opposite the largest interior angle.

Here we have

The side included by <1 and <2 is 11 cm

The side included by ∠2 and ∠3 is 16 cm  

The side included by ∠3 and <1 is 14 cm  

As we can observe angle ∠4 is an obtuse angle

Hence, ∠4 is the largest angle among the 4 angles

As we know the shortest side is placed opposite the smallest interior angle and the longest side is opposite the largest interior angle.

From the given data,

The side included by ∠2 and ∠3 is the longest side

Hence, the largest angles among ∠1, ∠2, and ∠3 are the angles ∠1

Therefore,

The second largest of the four angles is angle ∠1

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The monthly payment is ​$453.125.

If the initial amount (also called as principal amount) is P, and the interest rate is R% annually, and it is left for T years for that simple interest, then the interest amount earned is given by:

[tex]I = \dfrac{P \times R \times T}{100}[/tex]

Given;

Amount Financed (P) = $15,000

Annual Percentage Rate (R) = 5%

Number of Payments per Year (N) = 12

Time in Years (T) = 3

Now, monthly payment;

=(p*i*(1+i)^n)/((1+i)^n-1)

=15000(0.05/12)(1+0.05/12)^36/(1+0.05/12)^36-1

= 72.5/0.16

=453.125

Therefore, the monthly payment by 5% interest will be $453.125.

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The intersection of the set of integers and multiples of 5 will be (5,15).

What exactly are integers?

In mathematics, integers are the collection of positive and negative numbers. Integers, like whole numbers, do not include the fractional element. Thus, integers are numbers that can be positive, negative, or zero, but not fractions. On integers, we can do all arithmetic operations such as addition, subtraction, multiplication, and division. Integers include numbers such as 1, 2, 5, 8, -9, -12, and so on. "Z" is the symbol for integers.

Now,

Odd Integers between 0 and 20=1,3,5,7,9,11,13,15,17,19

Multiples of 5 between 0 and 35=5,10,15,20,25,30,35

Then the common integers are=5,15

Hence,

          The intersection of the set of integers multiples of 5 will be (5,15).

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Answer:

x²-2x-3 using factorisation method

Step-by-step explanation:

x²-3x+1x-3

x(x-3) 1(x-3)

(x+1) (x-3)

x=-1,x=+3

Answer: The annual percentage yield (APY) takes into account both the interest rate and the frequency of compounding. It is a measure of the effective rate of return over a year, taking into account the effect of compounding.

In Oxana's account, with a 7.1% interest rate

Step-by-step explanation:

The solution for the equation is [tex]x=\frac{\pi }{2} +2\pi n[/tex] ,for any integer n.

An equation is the statement that asserts th equality of the two expressions with 'equal to' (=) sign.

cot [tex]\frac{x}{2} =1[/tex],

Taking inverse on both sides we get,

[tex]\frac{x}{2}[/tex] = [tex]cot^{-1} (1)[/tex]⇒[tex]\frac{x}{2}=\frac{\pi}{4}[/tex]

                ⇒[tex]x=\frac{\pi }{2}[/tex].

cot is positive in first and third quadrant. To find the second solution add π to it.

[tex]\frac{x}{2} =\pi +\frac{\pi }{2}[/tex]

[tex]x=\frac{5\pi }{2}[/tex]

to find period for cot([tex]\frac{x}{2}[/tex])  

The periodic function can be calculated by [tex]\frac{\pi }{|b|}[/tex], replace b with [tex]\frac{1}{2}[/tex]  

we get [tex]2\pi[/tex]. we get period for  cot([tex]\frac{x}{2}[/tex])  is [tex]2\pi[/tex] so values will repeat every

radians in both directions.

[tex]x=\frac{\pi }{2} +2\pi n, \frac{5\pi }{2} +2\pi n,[/tex] for any integer n.

consolidating we get,

[tex]x=\frac{\pi }{2} +2\pi n[/tex], for any integer.

Hence, the solution for the equation cot [tex]\frac{x}{2}[/tex]= 1 is [tex]x=\frac{\pi }{2} +2\pi n[/tex]

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the best answer i got for a is 5 days (1.5 hours a day)

b was 3 days (an hour a day) but I am not that sure but thats the best answer I could come up with

The initial investment was $4,176.69

Investment is the dedication of money to purchase of an asset to attain an increase in value over a period of time.

To find initially invested we can use the formula for continuously compounded interest to find the initial investment:

A = Pe^(rt)

Where:

So, solving for P:

P = A / e^(rt)

P = $10,775.05 / e^(0.07 * 17)

P = $10,775.05 / 2.56724943786794

P = $4,176.69

Therefore, the initial investment was $4,176.69.

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None of the equations in Problems 1 through 8 is exact, so they cannot be solved using the method of integration factors.

If an ODE is exact, it can be solved using the method of integration factors.

Here is the determination of whether each of the equations in Problems 1 through 8 is exact, and if it is, finding the solution:

1. (2x + 3) + ( 2y - 2)y' = 0

This equation is not exact.

2. (2x + 4y) + ( 2x - 2y)y' = 0

This equation is not exact.

3. (3x^2 - 2xy + 2) + ( 6y^2 - x^2 + 3) y' = 0

This equation is not exact.

4. dx/dt = bx + cy, dy/dt = ax - by

This system of ODEs is not exact.

5. dx/dt = bx - cy, dy/dt = ax + by

This system of ODEs is not exact.

6. (ye^(2x) -2e^(2x)x sin(2x)+2x)+(xe^(2x)y cos(2x) -3) y'=0

This equation is not exact.

7. (y/x+6x) +(In x - 2)y' = 0, x > 0

This equation is not exact.

8. (y^2)dy/dx + 0 = (x^2 + y^2)^(3/2)

This equation is not exact.

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Answer:

254in^2

Step-by-step explanation:

(SA)=2lw+2lh+2hw

L -> 9 W ->4 H -> 7

2(9)(4)+2(9)(7)+2(7)(4)

18*4+18*7+14*

72+126+56

198+56

254

Answer: A

Step-by-step explanation:

If there is a $5000 Unearned Revenue credit on the accounting worksheet, it means that the company has received payment for products or services that have not yet been delivered. Therefore, when the delivery is made, an adjusting entry is needed to reflect the change in the company's financial statements.

The adjusting entry would be a debit to the Unearned Revenue account for $5000 and a credit to the Revenue account for $5000, reflecting the delivery of the products and the recognition of revenue. The entry would be recorded in the Adjustment column of the worksheet.

Therefore, the correct answer is option A, a $5000 Adjustment debit.

Answer:

A. a $5000 Adjustment debit

Step-by-step explanation:

The delivery of products listed as a $5000 Unearned Revenue credit would be recorded on an accounting worksheet with a $5000 Adjustment debit.

Unearned revenue is a liability account that represents payments received from customers for goods or services that have not yet been provided. When the delivery is made and the products are provided, the unearned revenue is "earned" and the revenue is recognized in the accounting records.

To record the delivery and recognize the revenue earned, the company would make an adjusting entry to decrease the unearned revenue account and increase the revenue account. Since the unearned revenue account has a credit balance, the adjusting entry would require a debit to reduce the balance. This would result in a $5000 Adjustment debit to Unearned Revenue and a $5000 Adjustment credit to Revenue.

The required function f ∶ R² → R given by f(x, y) = eˣ does not have any local or global extrema.

A global extreme point, also known as a global maximum or minimum, is the highest or lowest value that a function attains over its entire domain. In other words, it is the maximum or minimum value of the function over the whole range of inputs, rather than just a specific interval or region.

Here,

The function f(x, y) = eˣ only depends on x and is not a function of y. This means that it does not have any extrema since its value can be arbitrarily close to zero or arbitrarily close to infinity for different x-values, and therefore it does not attain a maximum or minimum value. This holds both locally and globally.

Thus, the required function f ∶ R² → R given by f(x, y) = eˣ does not have any local or global extrema.

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