Question 4 of 6What is the graph of the solution to the following compound inequality?15 3x-2 <16A.-345-2 -1102345-10-9-8-7 -6679 108B.-2-5 -4 -3-101234-10 -9 -8 -7 -656789 10C.-2 -10-10 -9 -8 -7 -6 -5 -4 -312345678910D.-4-3-2 -1-10 -9 -8 -7 -6 -50123465789 10SUBMIT

Answer:

(-5,1)

Step-by-step explanation:

Coordinates of midpoint

= ((-7-3)/2, (-5+7)/2)

= (-10/2 , 2/2)

= (-5, 1)

Answer:

she worked 12 hours as a carpenter and 18 hours as a blacksmith

Step-by-step explanation:

Let x = the number of hours for which she worked as a carpenter

Let y = be the number of hours for which she worked as a blacksmith

Total numbers of hours she worked = 30

x + y = 30 (1)

she earned a total $690. She earned $20 dollar per hour as a carpenter and $25 per hour as a blacksmith.

20 x + 25 y = 690 (2)

Equation (2) - 20 × Equation (1),

We get,

5 y = 90 ⇒ y = 18

After substituting this value in equation (1),

We get, x = 12,

Hence, She worked as a carpenter = x hours = 12 hours

And, as a blacksmith = y hours = 18 hours

Answer:

A=hbb 2=10·2 2=10cm²

Answer:

Step-by-step explanation:

We are given that in 2016, the aquarium had x fish, and in 2019, there were x3 fish. We also know that in 2019, there were 512 fish. So we can set up the equation:

x3 = 512

To solve for x, we need to take the cube root of both sides of the equation:

x = ∛512

Using a calculator or simplifying by prime factorization, we can find that:

512 = 2^9

So,

∛512 = ∛2^9 = 2^3 = 8

Therefore, there were 8 fish in the aquarium in 2016.

Answer:

Let's use algebra to solve the problem. We can start by using the information given to set up an equation relating the number of fish in 2016 (x) to the number of fish in 2019 (x3):

x3 = 512

To solve for x, we can isolate x by dividing both sides of the equation by 3:

x = 512/3

x = 170.67 (rounded to two decimal places)

Therefore, there were approximately 170.67 fish in the aquarium in 2016. However, since we cannot have a fraction of a fish, we can round up to the nearest whole number to get the final answer:

x ≈ 171

So there were 171 fish in the aquarium in 2016.

Step-by-step explanation:

Answer:

vdnchdbcdbchdbnhdbnhvbdjvjdsvhdfnjvdjvjdv

Step-by-step explanation:

The sum of the perimeters of the squares formed through geometric sequence is; 80 + 40√2.

We want to find the sum of the perimeters of the squares formed;

From the diagram attached, we see that;

In ∆PCQ;

<PCQ = 90°,

Then PQ² = PC² + QC²

Since, P and Q are mid - points of square ABCD, then it means that;

PC = BP = 5

QC = DQ = 5

Thus , PQ = √(25 + 25) =

PQ = √50 = 5√2

Since; E,F,G and H are mid-points of square PQRS, then it means that;

EP = PF = (5√2)/2

FQ = QG = (5√2)/2

Now , In ∆ FQG

∠FQG = 90°

Then ,FG²= FQ²+GQ²

FG² = 5

Similarly , The process the continued indefinitely .

Sum of the perimeters of the squares formed = perimeter of square ABCD+ Perimeter of square PQRS + Perimeter of square PQRS+....+∞

= (4 × 10) + (4 × 5√2) + (4 × 5) +......+∞

= 40 + 20√2 + 20 + ...... + ∞

In this Infinite geometric series, we see that;

First term; a = 40

Common ratio; r = (√2)/2

Sum of infinite GP is given by the formula;

Sₙ = a/(1 - r)

Thus;

Sₙ = 40/(1 - ((√2)/2)

This can be simplified by rationalizing the denominator to get 80 + 40√2.

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The time and electricians it would take the remaining 3 electricians 8 hours to complete the job if one electrician does not report.

More people to do a task means less time it will take.

Less people to do a task means more time it will take.

Thus, they are inversely related.

If x men take y time for a work,

We can define a constant as "Manpower" needed for doing that specific work.

Let we define:

Manpower needed for a work = y + y + y + .. + y = Time per man × count of men

Manpower needed for a work = xy

We are given that;

Number of electricians= 4

Number of hours= 6

Now,

We can use the formula:

workers × time = work

where "workers" is the number of electricians, "time" is the number of hours they work, and "work" is the amount of work done.

In this case, we know that 4 electricians can complete the job in 6 hours. So:

4 × 6 = work

work = 24

This means that the total amount of work required to complete the job is 24 "units".

Now, if one electrician doesn't show up, we have only 3 electricians to do the work. Let's call the time it takes for the 3 electricians to complete the job "t".

So we have:

3 × t = 24

Dividing both sides by 3, we get:

t = 8

Therefore, by work and time answer will be 3 electricians and 8 hours.

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The constant differences between the consecutive terms are 2 (a); 2 (b), -3 (c), 7 (d), 1(e), and 6(f).

In math, the constant difference can be defined as the number that defines the pattern of a sequence of numbers. This means that number that should be added or subtracted to continue with the sequence.

Due to this, to determine the constant difference it is important to observe the pattern and find out the number that should be added. For example, if the sequence is 2, 4, 6, 8, there is a difference of 2 between each of the numbers and this is the constant difference.

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Answer:

x=17

Step-by-step explanation:

Since ABCD is a rectangle, ∠DAB=90.

∠DAB=∠DAC+∠BAC

90=(2x+4)+(3x+1)

90=5x+5

5x=85

x=17

The value of distance of two points shown in figure is,

⇒ d = √20

We have to given that;

Two points on graph are,

(0, 1) and (2, - 3)

Since, The distance between two points (x₁ , y₁) and (x₂, y₂) is,

⇒ d = √ (x₂ - x₁)² + (y₂ - y₁)²

Hence, The distance of two points shown in figure is,

⇒ d = √ (x₂ - x₁)² + (y₂ - y₁)²

⇒ d = √ (2 - 0)² + (-3 - 1)²

⇒ d = √4 + 16

⇒ d = √20

Thus, The value of distance of two points shown in figure is,

⇒ d = √20

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Remember that

In order to find a coterminal angle or angles of the given angle, simply add or subtract 360 degrees of the terminal angle as many times as possible

Part a

we have

65 degrees

so

65+360=425 degrees

65-360=-295 degrees

Part B

we have

-125 degrees

so

-125+360=235 degrees

-125-360=-485 degrees

Answer:

what are the numbers my g

Step-by-step explanation:

The difference in temperature is 45.9F.

It is required to find the difference in temperature is F.

The degree or intensity of heat present in a substance or object, especially as expressed according to a comparative scale and shown by a thermometer or perceived by touch.

Given that:

The temperature in a town is 32.1 during the day and -13.8 at night.

The difference is

32.1 - (-13.8)

= 32.1 + 13.8

= 45.9F

Hence, the difference in temperature is 45.9F.

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data is missing

The temperature in a town is 32.1 during the day and -13.8 at night.

The percentage that can be filled with $3 in 1990 is: 29.41%

To calculate percentage growth rate:

Beginning:

Calculate the difference (increase) between the two numbers you are comparing. after that:

Divide the increment by the original number and multiply the result by 100. Growth rate = increment / original number * 100.  

We are told that it cost $3 to fill a gas tank as at 1970.

Now, there was a percentage price increase of (78.8 - 23.1)% = 55.2% from 1970 to 1990. Thus:

Cost of a gallon in 1970 = $0.36

Thus, number of gallons bought with $3 = 3/0.36 = 8.33 gallons at full tank

Now, in 1990, the cost is $1.23 and as such:

Quantity that can be bought = 3/1.23 = 2.45 gallons

Percentage of tank filled = 2.45/8.33 * 100% = 29.41%

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Answer:

p = y+3

Step-by-step explanation:

Recall the definition of absolute value:

\(|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}\)

Then we have a few cases to consider:

• If \(4x-5\ge0\) and \(-x+5\ge0\), the equation simplifies to

\(4x-5 = -x+5 \implies 5x=10 \implies \boxed{x=2}\)

• If \(4x-5<0\) and \(-x+5\ge0\), we get

\(-(4x-5) = -x + 5 \implies -3x = 0 \implies \boxed{x=0}\)

• If \(4x-5\ge0\) and \(-x+5<0\), then

\(4x-5 = -(-x+5) \implies 3x=0 \implies x=0\)

but we already have this solution.

• If \(4x-5<0\) and \(-x+5<0\), then

\(-(4x-5) = -(-x+5) \implies -5x = -10 \implies x=2\)

which we also already found.

Answer:

Domain:  (-∞, 2] U (3, 6]
Range: (-∞ , 1] U (-1, 4]

Step-by-step explanation:

This is a piecewise function. It has a discontinuity at x=2 and x = 3

The domain of a piecewise function is the union of the domains of each piece of the function. Similarly the range for a piecewise function is the union of the ranges for each piece of the function

Looking at the graph, we can see two subdomains. When x ≤ 2 , we have one subdomain and when x > 3 there is another subdomain. We state this as
D = - ∞ < x  ≤ 2 or 3 < x ≤ 6
In interval notation we represent this as

D =  (-∞, 2] U (3, 6]

The nature of the brackets is important a [ or ] bracket means that the value to which it is attached is included whereas an open bracket, (    or     ) means that value is not included

For example,  (-∞, 2] means that everything below x = 2 to -∞ is part of the domain but it does not include -∞; since at -∞, a function is not defined. However it does include 2 in the domain as can be seen by the closed bracket on the right

The other subdomain is for x > 3 (note: a filled circle means the value is included, an empty circle denotes the value is excluded)

The second subdomain is
3 < x ≤ 6
In interval notation, it is (3, 6]

So the total domain is the union of the two intervals

The range is also discontinuous and there are two distinct ranges. The range is nothing but all the set of values generated by the function. So we look at all possible values for y that the function produces

The Range of this function is  -∞ < f(x) ≤ 1 or -1 < f(x) ≤ -4
In interval notation this is
R = (-∞ , 1] U (-1, 4]

1.Bricks and timber purchased for construction. (Debit: Bricks - Rs 60,000, Debit: Timber - Rs 35,000, Credit: Bank - Rs 95,000)

2.Transfer of Rs 13,000 to fixed deposit account. (Debit: Fixed Deposit - Rs 13,000, Credit: Current Account - Rs 13,000)

3.Appointment of Mr. S.N. Rao as Accountant. (Debit: Salary Expense - Rs 300, Debit: Security Deposit - Rs 1,000, Credit: Accountant - Rs 300)

4.Goods sold to Shruti with discounts. (Debit: Accounts Receivable - Shruti - Rs 80,000, Credit: Sales - Rs 80,000)

5.Goods purchased from Richa with discounts. (Debit: Purchases - Rs 60,000, Credit: Accounts Payable - Richa - Rs 60,000)

6.Goods sold to Shilpa with discounts and received payment. (Debit: Accounts Receivable - Shilpa - Rs 50,000, Credit: Sales - Rs 50,000)

7.Goods sold to Garima with discounts and received partial payment. (Debit: Accounts Receivable - Garima - Rs 1,00,000, Credit: Sales - Rs 1,00,000)

8.Purchase of land with additional charges. (Debit: Land - Rs 2,00,000, Debit: Brokerage Expense - Rs 2,000, Debit: Registration Charges - Rs 15,000, Credit: Bank - Rs 2,17,000)

9.Proprietor took goods and cash for personal use. (Debit: Proprietor's Drawings - Rs 65,000, Credit: Goods - Rs 25,000, Credit: Cash - Rs 40,000)

10.Goods sold to Charu with profit and discount. (Debit: Accounts Receivable - Charu - Rs 1,20,000, Credit: Sales - Rs 1,20,000)

11.Rent paid for the building. (Debit: Rent Expense - Rs 60,000, Credit: Bank - Rs 60,000)

12.Goods sold to Sunil with profit and discount. (Debit: Accounts Receivable - Sunil - Rs 24,000, Credit: Sales - Rs 24,000)

13.Purchased goods from Nanda and supplied to Helen. (Debit: Purchases - Rs 1,000, Debit: Accounts Payable - Nanda - Rs 1,000, Credit: Accounts Receivable - Helen - Rs 1,300, Credit: Sales - Rs 1,300)

14.Purchased goods from Rohit and Sons and supplied to Madan. (Debit: Purchases - Rs 2,700, Credit: Accounts Payable - Rohit and Sons - Rs 2,700, Debit: Accounts Receivable - Madan - Rs 3,000, Credit: Sales - Rs 3,000)

Here are the journal entries for the given transactions:

1. Bricks and timber purchased for construction:

Debit: Bricks (Asset) - Rs 60,000

Debit: Timber (Asset) - Rs 35,000

Credit: Bank (Liability) - Rs 95,000

2. Transfer to fixed deposit account:

Debit: Fixed Deposit (Asset) - Rs 13,000

Credit: Current Account (Asset) - Rs 13,000

3. Appointment of Mr. S.N. Rao as Accountant:

Debit: Salary Expense (Expense) - Rs 300

Debit: Security Deposit (Asset) - Rs 1,000

Credit: Accountant (Liability) - Rs 300

4. Goods sold to Shruti:

Debit: Accounts Receivable - Shruti (Asset) - Rs 80,000

Credit: Sales (Income) - Rs 80,000

5. Goods purchased from Richa:

Debit: Purchases (Expense) - Rs 60,000

Credit: Accounts Payable - Richa (Liability) - Rs 60,000

6. Goods sold to Shilpa:

Debit: Accounts Receivable - Shilpa (Asset) - Rs 50,000

Credit: Sales (Income) - Rs 50,000

7. Goods sold to Garima:

Debit: Accounts Receivable - Garima (Asset) - Rs 1,00,000

Credit: Sales (Income) - Rs 1,00,000

8.Purchase of land:

Debit: Land (Asset) - Rs 2,00,000

Debit: Brokerage Expense (Expense) - Rs 2,000

Debit: Registration Charges (Expense) - Rs 15,000

Credit: Bank (Liability) - Rs 2,17,000

9. Goods and cash taken away by proprietor:

Debit: Proprietor's Drawings (Equity) - Rs 65,000

Credit: Goods (Asset) - Rs 25,000

Credit: Cash (Asset) - Rs 40,000

10. Goods sold to Charu:

Debit: Accounts Receivable - Charu (Asset) - Rs 1,20,000

Credit: Sales (Income) - Rs 1,20,000

Credit: Cost of Goods Sold (Expense) - Rs 80,000

Credit: Profit on Sales (Income) - Rs 40,000

11. Rent paid for the building:

Debit: Rent Expense (Expense) - Rs 60,000

Credit: Bank (Liability) - Rs 60,000

12. Goods sold to Sunil:

Debit: Accounts Receivable - Sunil (Asset) - Rs 24,000

Credit: Sales (Income) - Rs 24,000

Credit: Cost of Goods Sold (Expense) - Rs 20,000

Credit: Profit on Sales (Income) - Rs 4,000

13. Goods purchased from Nanda and supplied to Helen:

Debit: Purchases (Expense) - Rs 1,000

Debit: Accounts Payable - Nanda (Liability) - Rs 1,000

Credit: Accounts Receivable - Helen (Asset) - Rs 1,300

Credit: Sales (Income) - Rs 1,300

14. Goods received from Rohit and Sons and supplied to Madan:

Debit: Purchases (Expense) - Rs 2,700 (after 10% trade discount)

Credit: Accounts Payable - Rohit and Sons (Liability) - Rs 2,700

Debit: Accounts Receivable - Madan (Asset) - Rs 3,000

Credit: Sales (Income) - Rs 3,000

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Answer:

Step-by-step explanation:

First, we need to find the difference between the price that he sold the shares for and the price that he bought them at: (67.68-60.65) = 7.03

That means that there was a gain of $7.03 per share for Archie.

That being said, since Archie bought 100 shares, we can multiply that number by 100 to find the total gain from selling the shares:

(7.03x100)= 703

The answer then is:

Archie gained $703 from selling all of his shares.

Applying the inscribed angle theorem, the value of x in the given circle is: x = 35.

In the circle given, angle NWT is an inscribed angle in the given diagram and arc NT is the intercepted arc. According to the inscribed angle we have the following equation which we would use to solve for the value of x:

m<NWT = 1/2(measure of arc NT)

Plug in the values:

54 = 1/2(x + 2x + 3)

2(54) = x + 2x + 3

108 = 3x + 3

Solve for x:

108 - 3 = 3x

105 = 3x

105/3 = x

x = 35

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Answer:

Step-by-step explanation:

On a map, the distance between two towns in 2.6 cm. The scale of the map is 1 cm:50

km. What is the actual distance between the two towns? Calculator Allowed

Help please

Answer: x=68.5

Step-by-step explanation: You multiply 13.7 and 5 and you should get 68.5 as x.

\(20^{\tfrac 52}\\\\=(20^5)^{\tfrac 12}\\\\=\left(20^4 \cdot 20 \right)^{\tfrac 12}\\\\=20^{\tfrac 42} \cdot 20^{\tfrac 12}\\\\=20^2 \sqrt{20}\\\\=400\sqrt{4 \times 5}\\\\=400 \times 2\sqrt 5\\\\=800\sqrt 5\)

Answer:

\(800\sqrt{5}\)

Step-by-step explanation:

Given: \(\large (20)^\text{$ \dfrac{5}{2} $}\)

Properties of Exponents:

Rational Exponent Property: \(\large x^\text{$ \dfrac{m}{n} $} = \large \text{$ \sqrt[n]{x^m} $}\)

Product of Powers Property:

1. Convert into a radical:

\(\sqrt[2]{20^5} \implies \sqrt{20^5}\)

2. Simplify the expression:

\(\sqrt{20^2\times20^2\times20^1}\\\\\implies \sqrt{20^4\times20}\\\\\implies 20^2\sqrt{20}\\\\\implies 20^2\sqrt{4\times5}\\\\\implies 20^2\times2\sqrt{5}\)

3. Evaluate the power:

\(20\times20\times2\sqrt{5}\\\\\implies 400\times2\sqrt{5}\)

4. Multiply:

\((400\times2)\sqrt{5}\\\\\implies800\sqrt{5}\)

Answer:

y=6x-6

Step-by-step explanation:

The slope equation is: y=mx+b

Keep in mind that m= slope and b= y-intercept

Since this problem gives us both of the things needed to complete the equation, we just need to plug it in.

m, (slope) = 6

b, (y-intercept) = -9

y= 6x-6

Answer:

1: 10+6 x 6 = 96cm

2: 14 + 8 x 5 = 110cm

3: 9 + 14 x 20 = 460cm

Step-by-step explanation:

Hope this helps.

Answer:

54, 486, 154

Step-by-step explanation:

x+9x+100+x= 694

11x= 694-100

11x=594

x= 54

54×9=486

54+100= 154

54+486+154=694.

The sequence is will always have even numbers, is hence proved.

Given, the quadratic sequence is: 44:52:64:80

Sequences with a term are known as quadratic sequences. They can be distinguished by the fact that the first differences between terms are not equal but the second differences between terms are. Utilizing the term sum in an arithmetic progression formula, the sum of even numbers formula is achieved. The equation is Sum of Even Numbers Formula = n(n+1), where n is the total number of terms in the series.

The formula is: Tₙ = 2n² ₊ 2n  ₊ 40

take 2 common

Tₙ = 2(n² ₊ n  ₊ 20)

in other words n² ₊ n  ₊ 20  = y

therefore , Tₙ = 2y

which by definition makes Tₙ an even number at all times.

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When we reflect a point across the x-axis,

When we reflect a point across the y-axis,

The coordinates of B are (-3,-4).

Therefore, the coordinates of B' would be (-3,4)

(-3,4)

Answer:

I think l

Step-by-step explanation:

first add 96 and4 then 2 I think

Answer:

(2,1)

Step-by-step explanation:

The line 4y - 2x = 0 can be written in slope-intercept form as y = 1/2x.

So any point that satisfies this equation will lie on the line. For example, the point (2,1) satisfies the equation:

4(1) - 2(2) = 0

So the point (2,1) is on the line.