In the figure, m∠1=(x+14) ° and m∠2=(3x) ° . (a) Write an equation to find x. Male sure you use an “ = " sign in your answer.(b) Find the degree measure of each angle. Use the figure given in the picture to answer the questions.

Answer: the other midpoint is on (-19, 5)

Step-by-step explanation:

The slope of the line is -15, 4. And if the midpoint is (0,-4) then all you have to do is repeat the slope again and you’ll get the answer!

Answer:

1&2 are -2x^(3)+3x^(2)+x or Factored is −-x(2x^(2)-3x-1)

3&4 are 2x^(4)-3x^(3)-x^(2) or Factored is x^(2)(2x^(2)-3x-1)

5 is Yes

Step-by-step explanation:

1. (x+1*2x^(2)-x)(-1)

2x^(3)-x^(2)+2x^(2)-x(-1)

2x^(3)-3x^(2)-x(-1)

-2x^(3)+3x^(2)+x

−-x(2x^(2)-3x-1)

2. (2x^(2)-x*x+1*2x)(-1)

2x^(3)-x^(2)+2x^(2)-x(-1)

2x^(3)-3x^(2)-x(-1)

-2x^(3)+3x^(2)+x

−-x(2x^(2)-3x-1)

3. (x+1*2x^(2)-x)(x)

2x^(3)-x^(2)+2x^(2)-x(x)

2x^(3)-3x^(2)-x(x)

2x^(4)-3x^(3)-x^(2)

x^(2)(2x^(2)-3x-1)

4. (2x^(2)-x*x+1*2x)(x)

2x^(3)-x^(2)+2x^(2)-x(x)

2x^(3)-3x^(2)-x(x)

2x^(4)-3x^(3)-x^(2)

x^(2)(2x^(2)-3x-1)

5. (2x^(2)-x*x+1*2x)(x)=(x+1*2x^(2)-x)(x)

(2x^(2)-x*x+1*2x)(x)-2x^(2)-x=(x+1*2x^(2)-x)(x)-2x^(2)-x

(x+1)(x)=(x+1)(x)

(x+1)(x)-(x+1)=(x+1)(x)-(x+1)

X=X

Therefore, the interquartile range for the given data set is 27.5.

The interquartile range (IQR) is a measure of variability or spread of a set of data. It is calculated as the difference between the upper quartile (Q3) and the lower quartile (Q1) of a dataset. The quartiles divide the dataset into four equal parts, with each part containing an equal number of data points. The lower quartile (Q1) is the median of the lower half of the dataset, and the upper quartile (Q3) is the median of the upper half of the dataset.

Here,

To find the interquartile range (IQR), we first need to find the median of the data set. The median is the middle value when the data set is arranged in order. To arrange the data set in order, we have:

6, 12, 14, 15, 15, 20, 35, 48, 87

The median is the middle value, which is 15.

Next, we need to find the median of the lower half of the data set (also called the first quartile or Q1). To do this, we take the median of the values below 15:

6, 12, 14, 15, 15

The median of this set is 14.

Similarly, we need to find the median of the upper half of the data set (also called the third quartile or Q3). To do this, we take the median of the values above 15:

20, 35, 48, 87

The median of this set is (35+48)/2 = 41.5.

Finally, the interquartile range is the difference between Q3 and Q1:

IQR = Q3 - Q1

= 41.5 - 14

= 27.5

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Answer: 51.34

Step-by-step explanation:

you need to use SOH CAH TOA

here they give opp and adj

Use tan

tan∅=opp/adj=10/8   now use that tan^(-1) (10/8)  in calculator

x=51.34

Answer:

0.04545454545

Step-by-step explanation:

Divide 1 by 22

Answer:

0.0454545454545454545

Step-by-step explanation:

(1/2 times 1/11)

Answer:

  C, E, D

Step-by-step explanation:

A "unit rate" is a rate that has 1 unit in the denominator. Here, the rates are all given in terms of the number of apples in 1 basket. (The 1 basket is the 1 unit.)

When rates are given with some number of units other than 1 in the denominator, divide the numerator by the denominator to find the unit rate.

Here, the unit rates are all more than one apple in each basket, so the number of apples must be a multiple of the number of baskets. Phrases "A" and "B" both have more baskets than apples, so will not be used.

  C: (304 apples)/(8 baskets) = (304/8) apples/basket = 38 apples/basket

  D: (294 apples)/(7 baskets) = (294/7) apples/basket = 42 apples/basket

  E: (405 apples)/(9 baskets) = (405/9) apples/basket = 45 apples/basket

  (38 apples)/(1 basket) = C: 38 apples/basket

  (45 apples)/(1 basket) = E: 45 apples/basket

  (42 apples)/(1 basket) = D: 42 apples/basket

A). The function is symmetric about the line x = 1 and continuous.

For 0 x 2, the function is positive. The highest value for the function is

(1, 1). The value of f(x) increases as x approaches positive infinity.

Each input value is given a distinct output value by a rule known as a function.

Functions can be shown using graphs, tables, mathematical notation, and other techniques.

The function is positive in the range 0 x 2, therefore we can limit the curve to that region. As a result, the curve may increase quickly as x moves away from 2. The produced graph might look like this:  

B). The function is continuous and symmetrical about the line x = 2. For the function, the bare minimum is (2, 3). As x approaches positive or negative infinity, f(x) approaches infinity.

Similar to how we can design a symmetric curve with a minimum point at x = 2 because the function is symmetric around that value. (2, 3). When x gets close to positive or negative infinity, the function moves towards infinity.  

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Answer:

a) 293 units

b) 5795p

Step-by-step explanation:

First, determine how many units used

Take the new reading and subtract the old reading

6538 - 6245

293 units

The first 200 is at 22p

200 * 22 p =4400p

293 -200 = 93 units at the remaining unit price.

93 * 15 p =1395p

Add the two totals together

4400p + 1395p =5795p

Answer:

level of water in a reservoir each....

Step-by-step explanation:

The scatter plot

A scatter plot is like a scatter chart or graph that uses the dots for representing the values of two different numeric variables. The horizontal value indicates the individual point data. Thus is used for observing the realtisohips between the variables.

This the answer is level of the water in a dam each day and date during 1 month.

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the assumption here is that both triangles are similar, if that's so, then we can use proportions to get the sides.

\(\stackrel{\textit{assuming both triangles are similar}}{\cfrac{x}{10}=\cfrac{15}{6x}\implies \cfrac{x}{10}=\cfrac{5}{2x}}\implies 2x^2=50 \\\\\\ x^2=\cfrac{50}{2}\implies x^2=25\implies x=\sqrt{25}\implies \boxed{x=5}\)

For the equation 75x+30y=1500,  x is (100 - 2y) / 5 and y is (100 - 5x) / 2.

An equation is a mathematical statement that uses symbols and numbers to express a relationship between two or more variables. It states that two expressions are equal to each other.

We can solve for x and y in the equation 75x + 30y = 1500 by using algebraic manipulation.

First, we can simplify the equation by dividing both sides by the greatest common factor of 75 and 30, which is 15:

(75x + 30y) / 15 = 1500 / 15

Simplifying the left side gives:

5x + 2y = 100

Next, we can isolate one of the variables. Let's isolate y by subtracting 5x from both sides:

5x + 2y - 5x = 100 - 5x

Simplifying the left side gives:

2y = 100 - 5x

by dividing both sides by 2:

y = (100 - 5x) / 2

This is the equation for y in terms of x. We can use this equation to solve for y given a value of x, or we can substitute it back into the original equation to solve for x in terms of y.

Let's solve for x in terms of y by isolating x in the equation 5x + 2y = 100:

5x = 100 - 2y

Dividing both sides by 5 gives:

x = (100 - 2y) / 5

This is the equation for x in terms of y. We can use this equation to solve for x given a value of y, or we can substitute it back into the original equation to solve for y in terms of x.

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The coefficient of kinetic friction between the oil-covered road

and the car tires. is 0.51

To solve this question, we would consider one of the equations of motion.

v² = u² + 2as, where

v = final velocity, which is also equal to 0

u = initial velocity, which is 20m/s

s = skid marks, which is 40 meters long

a = acceleration, which is unknown

0² = 20² + 2*a*40

0 = 400 + 80a

80a = -400

a = -400/80

a = -5 m/s

Also, it is known that acceleration a = (mu)g, where mu is the coefficient of kinetic friction and g is the acceleration due to gravity.

-5 = mu * 9.8

mu = -5/9.8

mu = 0.51

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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:

10 * 10 and you find the square, add (6 * 4) : 2 (triangle) and you have the total area

10 * 10 = 100

(6 * 4) : 2 = 12

100 + 12 = 112in²

Answer:

Step-by-step explanation:

To solve simultaneous equations graphically, you just sketch the lines on a graph and look for the point of intersection.

To sketch x + 2y = 63, just find the x and y intercepts. At the x-intercept, the y co-ordinate is always zero, so sub this in for y and solve for x:

⇒ x + 2(0) = 63

⇒ x = 63

So the x-intercept is (63, 0)

We can do the opposite for the y-intercept, as the x co-ordinate is always 0:

⇒ 0 + 2y = 63

⇒ 2y = 63

⇒ y = 31.5

So the y-intercept is (0, 31.5)

Just repeat for the second line, mark the intercepts on a graph, join them, and then look for the co-ordinates of where the lines intersect. These will be the values for x and y.

For substitution, we rearrange the subject for one equation, then sub its value into the other one.

So we can rearrange x - y = 4 to get x = y + 4. Now we sub the value of x here into the other equation:

⇒ (y + 4) + 2y = 63

Then just solve for y and plug it into either equation and find x.

For elimination, we can either add or subtract the two equations to eliminate a term. Here, we can subtract the two equations, to get rid of x:

⇒ x + 2y = 63

 - x  -  y  =  4

x - x is 0, so it is eliminated, 2y - - y is 2y + y which is 3y, and 63 -4 is 59.

So we're left with 3y = 59, which is easy to solve and then we plug it back in to solve for x.

Answer:

The illustration is found in the attachment below.

Note: The correct order of the question is found below:

Learning Task 1: Illustrate the following fractions on your notebook.

a.)2/3 b.)7/9 c.)10/12 d.)8/12 e.)4/5 f.) 8/6 g.)9/4 h.) 28/10

Step-by-step explanation:

a) Draw a rectangle of sides 3 cm by 1 cm. Divide the rectangle into three equal parts. Shade two out of the three parts to give the fraction 2/3

b)  Draw a rectangle of sides 9 cm by 1 cm. Divide the rectangle into nine equal parts. Shade seven out of the nine parts to give the fraction 7/9

c)  Draw a rectangle of sides 12 cm by 1 cm. Divide the rectangle into twelve equal parts. Shade ten out of the twelve parts to give the fraction 10/12

d)  Draw a rectangle of sides 12 cm by 1 cm. Divide the rectangle into twelve equal parts. Shade eight out of the twelve parts to give the fraction 8/12

e)  Draw a rectangle of sides 5 cm by 1 cm. Divide the rectangle into five equal parts. Shade four out of the five parts to give the fraction 4/5

f)  Draw two rectangle of sides 6 cm by 1 cm. Divide the rectangles into six equal parts each. Shade all the six parts of the first rectangle completely. Then, shade two out of the six parts of the second rectangle to give the fraction 8/6

g) Draw three rectangles of side 4 cm by 1 cm. Divide each of the three rectangles into four parts each. Shade completely the four parts of the first and second rectangle. Then, shade one part out of the four parts of the third rectangle to give the fraction 9/4

h) Draw three rectangles of side 10 cm by 1 cm. Divide each of the three rectangles into ten parts each. Shade completely the ten parts of the first and second rectangle. Then, shade eight parts out of the ten parts of the third rectangle to give the fraction 28/10

Answer: Perimeter: 68 Area: 289

Step-by-step explanation:

17 by 17 is basically a height of 17 and a length of 17, so if the shape is 2d, than the perimeter is (17+17)*2, and the area is 17*17.

Answer:

0.4054 = 40.54% probability of selecting a black sock on the second draw given that a black sock was selected on the first draw

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

\(P(B|A) = \frac{P(A \cap B)}{P(A)}\)

In which

P(B|A) is the probability of event B happening, given that A happened.

\(P(A \cap B)\) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Black sock on the first draw.

Event B: Black sock on the second draw.

The probability of selecting a black sock on the first draw is 8/19.

This means that \(P(A) = \frac{8}{19}\)

Black socks on both draws:

The probability of selecting two black socks is 120/703

This means that \(P(A \cap B) = \frac{120}{703}\)

What is the probability of selecting a black sock on the second draw given that a black sock was selected on the first draw?

\(P(B|A) = \frac{P(A \cap B)}{P(A)}\)

\(P(B|A) = \frac{\frac{120}{703}}{\frac{8}{19}}\)

\(P(B|A) = \frac{120}{703}*\frac{19}{8}\)

\(P(B|A) = \frac{120*19}{703*8}\)

\(P(B|A) = 0.4054\)

0.4054 = 40.54% probability of selecting a black sock on the second draw given that a black sock was selected on the first draw

Answer:

6/6 x 3/5, 1/4 x 3/5, 1 2/5 x 3/5, 2 x 3/5

Hope this helps

Step-by-step explanation:

Answer:

3x = 54 !

Step-by-step explanation:

3x = (72 - x)

Solve for x.

x = 18

Plug it in.

3(18) = 54 degrees

Hope this helps! Xx

Answer:

568 candies

Step-by-step explanation:

16 oz/ 1.69 oz.= 9.47 x 60 candies= 568.2= 568 candies.

The ways to pack are

This is further explained below.

Generally, The science, art, and technology of packaging involves confining or safeguarding goods for distribution, storage, sale, and usage. The process of creating, analyzing, and designing packages is sometimes referred to as packaging.

In conclusion,

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Tasha paid an initial fee of $25 plus $30 per hour for six hours of instruction, resulting in a total cost of $205.

Let's assume the initial fee charged by the skiing instructor is represented by 'x' dollars. Since Tasha paid $205 for six hours of instruction, we can set up an equation to represent the total cost:

Total cost = Initial fee + (Hourly rate * Number of hours)

205 = x + (30 * 6)

205 = x + 180

To isolate the initial fee 'x', we subtract 180 from both sides of the equation:

205 - 180 = x

25 = x

Therefore, the initial fee charged by the skiing instructor is $25.

Now, let's calculate the total cost for Tasha's six hours of instruction using this information:

Total cost = Initial fee + (Hourly rate * Number of hours)

Total cost = $25 + (30 * 6)

Total cost = $25 + $180

Total cost = $205

We can see that the calculated total cost matches the amount Tasha paid, confirming the accuracy of the calculation.

To summarize, Tasha paid an initial fee of $25 plus $30 per hour for six hours of instruction, resulting in a total cost of $205. The initial fee covers the fixed cost, while the hourly rate accounts for the variable cost based on the number of hours of instruction received

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Answer: The usual dose for an 18-month-old weighing 21 pounds is 48 mg of ibuprofen.

Step-by-step explanation: To find the usual dose of ibuprofen for a child, we need to follow these steps:

Therefore, the usual dose for an 18-month-old weighing 21 pounds is 48 mg of ibuprofen. Hope this helps, and have a great day! =)