Mindy is saving money. She started with $0. After 6 weeks, she had $90 saved.

Mindy is not sure exactly how much money she saved each week. She assumes that she saved money at a constant rate from when she started saving money through week 6

There is a proportional relationship between the number of pixels and the area of the square.

Two sets of numbers are proportional or directly proportional if their corresponding elements have a constant ratio, which is said the coefficient of proportionality or proportionality constant. Two sets are inversely proportional if corresponding elements have a constant product, also defined as the coefficient of proportionality.

According to the question,

\(\frac{number of pixels}{square inches of screen} = constant value\)

\(\frac{31.104}{6} = \frac{373.248}{72} = \frac{544.320}{105} = 5.184\)

So, the relationship is proportional.

The equation is :

a = 5184p

where p represents the number of pixels and a is the area of the screen in square inches.

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Answer:

McKinley, Logan, Pico de Orizaba

Step-by-step explanation:

The amount of interest she will earn at the end of 3 years would be = $375

The principal amount of money saved by Chantal (p)= $5000

The interest rate that the amount generates (r)= 2.5%

The total time for the savings(t) = 3 years.

The simple interest generated S.I = P×T×R/100

= 5000×3×2.5/100

= 37500/100

= $375

Therefore the amount of money she will earn as interest after three years of savings would be = $375.

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Answer:

Step-by-step explanation:

Subtract equation 2 from equation 1

.7x+y-(-1.5x+y)=340-320

.7x+y+1.5x-y=20

2.2x=20

x=100/11, then use this in either original equation to find y.

.7(100/11)+y=340

y=3670/11, so the balance point is

(100/11, 3670/11)

We need to determine the number of integers greater than 1000 that can be formed using the digits 0, 2, 3, and 5 without repeating any digit. To form integers greater than 1000, the thousands place must be occupied by one of the digits 2, 3, or 5. The hundreds, tens, and units places can be filled with any of the remaining digits.

Case 1: Thousands place digit is 2

In this case, we have three choices for the thousands place (2, 3, or 5). After selecting the thousands place digit, the remaining three digits can be arranged in the hundreds, tens, and units places in 3! = 6 ways. Therefore, for this case, we have 3 * 6 = 18 integers greater than 1000.

Case 2: Thousands place digit is 3

Similarly, we have three choices for the thousands place (2, 3, or 5). After selecting the thousands place digit, the remaining three digits can be arranged in the hundreds, tens, and units places in 3! = 6 ways. Hence, for this case, we also have 3 * 6 = 18 integers greater than 1000.

Case 3: Thousands place digit is 5

Again, we have three choices for the thousands place (2, 3, or 5). After selecting the thousands place digit, the remaining three digits can be arranged in the hundreds, tens, and units places in 3! = 6 ways. Thus, for this case, we have 3 * 6 = 18 integers greater than 1000.In total, we have 18 + 18 + 18 = 54 integers greater than 1000 that can be formed using the digits 0, 2, 3, and 5 without repeating any digit.

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Scientific notation and engineering notation of 19,485,000 Scientific notation of

19,485,000= 1.9485 × 10⁷

Engineering notation of 19,485,000= 19.485 × 10⁶

b) Scientific notation and engineering notation of 834,7050Scientific notation of

834,7050= 8.34705 × 10⁶

Engineering notation of

834,7050= 83.4705 × 10⁴

c) Scientific notation and engineering notation of 5,835,00,000Scientific notation of 5,835,00,000= 5.835 × 10⁹

Engineering notation of 5,835,00,000= 58.35 × 10⁸d) Scientific notation and engineering notation of 0.0000053 Scientific notation of

0.0000053= 5.3 × 10⁻⁶

Engineering notation of

0.0000053= 53 × 10⁻⁸

e) Scientific notation and engineering notation of 0.00000081

Scientific notation of

0.00000081= 8.1 × 10⁻⁷

Engineering notation of

0.00000081= 81 × 10⁻⁹

f) Scientific notation and engineering notation of

0.0000012

Scientific notation of

0.0000012= 1.2 × 10⁻⁶

Engineering notation of

0.0000012= 12 × 10⁻⁸ \\

Scientific notation is used to represent numbers that are very large or very small. It is a method of expressing numbers in terms of a decimal number between 1 and 10 and a power of 10. Engineering notation is a version of scientific notation in which the power of ten is always a multiple of three. Both scientific and engineering notations can be used to represent large and small numbers in a more convenient form. Scientific notation is written in the form a × 10ⁿ, where 1 ≤ a < 10 and n is an integer.

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Answer:

-5 and 0

Step-by-step explanation:

7x+2

Step-by-step explanation:

add all the sides of the pentagon and make it equal to the perimeter of the square which is 4y

sides of the pentagon are

5x+3

2x+3

7x+4

9x-10

5x+8

28x+8=4y

7x+2 = y

7x+2 = answer

gathmath

gathmathier6961

another way :

in a standardized test you may want to use real numbers

make x equal a prime number like 3

so then the sides are

18

9

25

17

23

=

92

then divide by 4

23

then divide by 3

which is 7 remainder of 2

which is

7x+2

y = 5x + 3

-1 = 5(2) + 3

-1 = 13 (wrong)

y = -2x - 4

-1 = -2(2) - 4

-1 = -8 (wrong)

It will take approximately 3.96 years (or approximately 3 years and 11 months) for Vincent's savings to reach $9,000.

To solve this problem, we can use the formula for the future value of an ordinary annuity:

FV = P * [(1 + r)^n - 1] / r

Where:

FV = Future Value

P = Payment (deposit amount)

r = Interest rate per compounding period

n = Number of compounding periods

In this case, Vincent deposits $165 at the end of each quarter, so P = $165. The annual interest rate is 6%, compounded quarterly, so the interest rate per compounding period, r, would be 6% divided by 4 (since there are 4 quarters in a year), which is 0.06/4 = 0.015.

We want to find the number of years, n, it will take for Vincent's savings to reach $9,000. Let's substitute the given values into the formula and solve for n:

9,000 = 165 * [(1 + 0.015)^n - 1] / 0.015

To solve this equation, we can use algebraic techniques or a financial calculator. Solving the equation, we find that n is approximately 15.83.

Since n represents the number of quarters, we need to convert it to years by dividing it by 4 (since there are 4 quarters in a year):

n (in years) = 15.83 / 4 ≈ 3.96

Therefore, it will take approximately 3.96 years (or approximately 3 years and 11 months) for Vincent's savings to reach $9,000.

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These transformations result in the graph of g(x), which is a horizontally reflected, horizontally compressed, horizontally shifted, and vertically shifted version of the parent function f(x) = log(x).

To achieve the graph of g(x) = log("-2x" - 4) - 1 from the parent function f(x) = log(x), several transformations are applied.

First, a horizontal reflection occurs due to the negative coefficient in front of the x term, "-2x". This reflection flips the graph of f(x) across the y-axis.

Next, a horizontal compression takes place due to the coefficient of 2 in "-2x". This compression squeezes the graph horizontally, making it narrower compared to the parent function.

Then, a horizontal shift to the right occurs by 4 units due to the constant term -4. This shift moves the graph horizontally, shifting it to the right.

Lastly, a vertical shift downward by 1 unit takes place due to the constant term -1. This shift moves the entire graph vertically, shifting it downward.

These transformations result in the graph of g(x), which is a horizontally reflected, horizontally compressed, horizontally shifted, and vertically shifted version of the parent function f(x) = log(x).

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Answer:

Step-by-step explanation:

We get the Polynomials  quotient and remainder as 2x+7 and 5 respectively.

An expression that consists of variables, constants, and exponents that is combined using mathematical operations like addition, subtraction, multiplication, and division is referred to as a polynomial.

A polynomial is an equation made up of coefficients and indeterminates that uses only the addition, subtraction, multiplication, and powers of positive-integer variables.

Polynomial long division is an extended variant of the well-known mathematical operation known as long division. It is an algorithm for dividing a polynomial by another polynomial of the same or lower degree. Due to the fact that it breaks down a more difficult division issue into simpler ones, it may be completed quickly by hand.

The polynomials are

2X^2 - X +16

and

x - 3

Dividing  by 2X^2 - X +16 BY x-3 = 2x + 7

We get the quotiant and remainder as 2x+7 and 5 respectively.

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Accounts receivable (Art) $1,078Sales Discount    $22Cash $1,100

Journal entry is a process of recording every transaction in a business in the order they take place. In the case of Cabela's, the following merchandising transactions were recorded:1. Purchased merchandise from Boden Company for $6,000 under credit terms of 1/15, n/30, FOB shipping point, invoice dated July 1The cost of merchandise purchased from Boden Company is debited while Accounts payable (Boden Company) is credited. The cost of merchandise is calculated as $6,000 * (1-1%) = $5,940 while the remaining $60 is the discount received.

Cost of merchandise   $5,940Accounts payable (Boden)     $5,9402. Sold merchandise to Creek Company for $900 under credit terms of 2/10, n/60, FOB shipping point, invoice dated July 2.The sales revenue is credited while the cost of goods sold is debited with the cost of the merchandise. This is because the cost of merchandise sold is an expense and expenses are debited.

Sales revenue $900Cost of goods sold $500Merchandise inventory $5003. Paid $125 cash for freight charges on the purchase of July 1The Freight Out expense is debited while Cash is credited.

Freight out expense $125Cash $1254. Sold merchandise that had cost $1,300 for $1,700 cashThe cash received for the sale is debited while the cost of goods sold is credited with the cost of the merchandise. This is because the cost of merchandise sold is an expense and expenses are debited.

Cash $1,700Cost of goods sold $1,300Merchandise inventory $1,3005. Purchased merchandise from Leight Company for $2,200 under credit terms of 2/15, n/60, FOB destination, invoice dated July 9The cost of merchandise purchased is debited while Accounts payable (Leight Company) is credited. The cost of merchandise is calculated as $2,200 * (1-2%) = $2,156 while the remaining $44 is the discount received.

Cost of merchandise   $2,156Accounts payable (Leight)     $2,1566. Returned $200 of merchandise purchased on July 9 from Leight Company and debited its account payable for that amount.The Accounts payable is debited with the amount of merchandise returned while the Merchandise inventory account is credited with the cost of the returned merchandise.Accounts payable (Leight)   $200Merchandise inventory $2007. Received the balance due from Creek Company for the invoice dated July 2, net of the discountThe accounts receivable account is debited with the net amount received while the sales discount account is credited with the discount received.

Accounts receivable (Creek) $882Sales discount     $18Cash $9008. Sold merchandise that cost $800 to Art Company for $1,200 under credit terms of 2/15, n/60, FOB shipping point, invoice dated July 19The sales revenue is credited while the cost of goods sold is debited with the cost of the merchandise.

This is because the cost of merchandise sold is an expense and expenses are debited.Sales revenue $1,200Cost of goods sold $800Merchandise inventory $8009. Gave a price reduction (allowance) of $100 to Art Company for merchandise sold on July 19 and credited Art's accounts receivable for that amount.

The allowance account is debited while Accounts receivable (Art Company) is credited.Allowance    $100Accounts receivable (Art) $10010. Paid Leight Company the balance due, net of discountAccounts payable (Leight) is debited with the net amount paid while Cash is credited.Accounts payable (Leight)   $2,128Cash $2,12811.

Received the balance due from Art Company for the invoice dated July 19, net of discount The accounts receivable account is debited with the net amount received while the sales discount account is credited with the discount received.

Accounts receivable (Art) $1,078Sales discount    $22Cash $1,100

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Answer:

a. 60%

b. 37%

c. 3%

Step-by-step explanation:

a. 72/ 120 then multiply by 100

b. 120- 72 + 4 = 44/ 120 x 100

c. 4/120 x 100

Answer:

a) 60%.      b) 37%.      c) 3%

The remainder obtained when the quotient remainder theorem is applied to n with divisor d is 0.

According to the quotient remainder theorem, for any integer n and positive integer d, there exist unique integers q and r such that n = qd + r and 0 ≤ r < d. In other words, n can be expressed as the product of d and some integer q plus a remainder r, where the remainder is always less than the divisor.

If d | n, this means that n is divisible by d with no remainder. Therefore, the quotient q will be an integer and the remainder r will be 0.

For example, if d = 3 and n = 12, then d | n because 12 is divisible by 3 with no remainder. The quotient remainder theorem can be applied as follows:

n = qd + r
12 = 3(4) + 0

The quotient q is 4 and the remainder r is 0. Therefore, the remainder obtained when the quotient remainder theorem is applied to n with divisor d is 0.

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The mean number of garage stalls per home in the sample of 200 homes is 1.09.

The mean is a measure of central tendency in statistics that represents the average value of a set of numerical data. It is calculated by summing up all the values and dividing by the total number of values.

According to the given information:

To find the mean number of garage stalls per home in the sample of 200 homes, we need to calculate the expected value or the average value of X, which is the number of garage stalls per home. We are given the probabilities of X taking the values 0, 1, and 2, which are P(X=0) = 0.06, P(X=1) = 0.45, and P(X=2) = 0.32.

The formula for calculating the expected value of X is:

E(X) = Σ [ x × P(X=x) ]

where Σ represents the sum of all values of x, and P(X=x) is the probability of X taking the value x.

Using this formula, we can calculate the expected value of X as follows:

\(E(X) = (0 * 0.06) + (1 * 0.45) + (2 * 0.32)\)

\(= 0 + 0.45 + 0.64\)

= 1.09

Therefore, the mean number of garage stalls per home in the sample of 200 homes is 1.09. Hence, the correct option is (a) 1.09.

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The mean number of garage stalls per home in the sample of 200 homes is 1.09.

What is mean?

The mean is a measure of central tendency in statistics that represents the average value of a set of numerical data. It is calculated by summing up all the values and dividing by the total number of values.

According to the given information:

To find the mean number of garage stalls per home in the sample of 200 homes, we need to calculate the expected value or the average value of X, which is the number of garage stalls per home. We are given the probabilities of X taking the values 0, 1, and 2, which are P(X=0) = 0.06, P(X=1) = 0.45, and P(X=2) = 0.32.

The formula for calculating the expected value of X is:

E(X) = Σ [ x × P(X=x) ]

where Σ represents the sum of all values of x, and P(X=x) is the probability of X taking the value x.

Using this formula, we can calculate the expected value of X as follows:

E(x) = (0.06*0)+(1*0.45)+(2*0.32)

= 1.09

Therefore, the mean number of garage stalls per home in the sample of 200 homes is 1.09. Hence, the correct option is (a) 1.09.

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Answer:

430

Step-by-step explanation:

Therefore, the answer is 430.

The equation that could be used to solve for the time t in hours that she drove before her stop is 80t + 100(8/3 - t) = 250.

An algebraic expression is the combination of numbers and variables in expressing and solving a particular mathematical question. An equation is the equality of expressions.

Let t = time she drove before her stop

If the total trip time is 3 hours including the stop, which took 20 minutes, then the time she drove after the stop is 3 - t - 1/3 or 8/3 - t.

If altogether she drove a distance of 250 km, then the sum of the products of the rate of speed and time each drive is equal to 250.

(80 km/h)(t) + (100 km/h)(8/3 - t) = 250

Simplify the equation.

80t + 100(8/3 - t) = 250

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9 block rhombuses would create 3 hexagons.

In order to increase the total number by three, a hexagon can be divided into a minimum of three rhombuses, each of which can be further divided into four rhombuses. With n being a positive integer, the result is 3n. The top two hexagons in this image are each divided into 8 and 10 rhombuses. The total number of rhombuses that can fit in a hexagon is not limited to multiples of three, given that the total number of rhombuses dividing a hexagon can be raised by three by quartering a single component rhombus. The amount might actually be any integer larger than 7.

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Answer:

Step-by-step explanation:

yes

Answer:

Step-by-step explanation:

if x = y, and y = z, you can write that as x = y = z, which means that x = z

Answer:

Angle = A

A7 = A4 (Alternate interior angles)

A14 = A11 (Vertically opposite angles)

A11 = A6(Corresponding angles)

A4 = A14 (Given)

From this, we can get A6 = A4

A6 and A4 are alternate angles and are equal. So c and d are parallel.

A7 + A10 = 180 (Linear pair of angles) - ( 1 )

A13 = A10(Alternate interior angles)

Putting A13 in place of A10 in ( 1 ),

A7 + A13 = 180

Hence proved

Ask doubts if any

Remember that if f and g are inverses of one another, then

f(g(x)) = g(f(x)) = x

1/2. Take a = 0 and b = 1 (or any non-zero number) so that

f(x) = x + 0/1   ⇒   f(x) = x

If g is to be an inverse of f, we need

g(f(x)) = g(x) = x

so that c = 1 and d = 0.

3. With f(x) = x + a/b and g(x) = cx - d, we have

g(f(x)) = g(x + a/b) = c (x + a/b) - d = cx + ac/b - d

and of course, with a,b,c,d as before, we get g(f(x)) = x.

4. This would be a very uninteresting graph for the example I've cooked up here, just containing the line y = x...

The correct answer is: c. Gain of £800,000 to Statement of Profit or loss.

Since Maris uses the fair value model, the gain from the increase in the fair value of the building is recognized in the Statement of Profit or Loss. In this case, the building's fair value increased from £10 million to £10.8 million, resulting in a gain of £800,000. Therefore, the gain of £800,000 should be recognized in the Statement of Profit or Loss.According to the fair value model, any gain or loss resulting from the change in fair value of the asset should be recognized in the financial statements. In this case, the increase in the fair value of the building is considered a gain.

Since the gain of £800,000 (the difference between the fair value of £10.8 million and the original purchase price of £10 million) is a result of the change in the asset's fair value, it should be recognized in the Statement of Profit or Loss. This gain represents the increase in the value of the building during the year.

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Answer:

2.5x+50

Step-by-step explanation:

From the drawing above :

• Sin (B) = opp/hyp = (b /c )= 55/73

• Cos (B) = adj / hyp =(a/c ) = 48 /73

• Tan(B) = opp/adj = (b/ a) = 55/48