i need help with this

Answer:

a) we get value of n: n=10.7 or n=1.2 when d=30

b) we get value of n: n=9.6 or n=2.3 when d=20

Step-by-step explanation:

a) solve for  n When  do= 30

Put d= 30 in the given equation:

\(d = n^2-12n + 43\)

\(30=n^2-12n+43\\n^2-12n+43-30=0\\n^2-12n+13=0\\\)

Now, we will find value of n by using quadratic formula: \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

We have a=1, b=-12 and c=13

\(n=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\n=\frac{-(-12)\pm\sqrt{(-12)^2-4(1)(13)}}{2(1)}\\n=\frac{12\pm\sqrt{144-52}}{2}\\n=\frac{12\pm\sqrt{92}}{2}\\n=\frac{12\pm9.59}{2}\\n=\frac{12+9.59}{2} , n=\frac{12-9.59}{2}\\n=10.7 , n=1.2\\\)

So, we get value of n: n=10.7 or n=1.2

b) solve for n  when  d=20

\(d=n^2-12n+43\\20=n^2-12n+43\\n^2-12n+43-20=0\\n^2-12n+23=0\\\)

Now, we will find value of n by using quadratic formula: \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

We have a=1, b=-12 and c=23

\(n=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\n=\frac{-(-12)\pm\sqrt{(-12)^2-4(1)(23)}}{2(1)}\\n=\frac{12\pm\sqrt{144-92}}{2}\\n=\frac{12\pm\sqrt{52}}{2}\\n=\frac{12\pm7.21}{2}\\n=\frac{12+7.21}{2} , n=\frac{12-7.21}{2}\\n=9.6 , n=2.3\\\)

So, we get value of n: n=9.6 or n=2.3

Answer:

8 more red counters

Step-by-step explanation:

It would make it 9:3, which, when simplified, is 3:1.

Answer:

Step-by-step explanation:

words algebraic symbols reason

two decreased by a number   2 - x   start with 2, make it smaller

a number decreased by two   x - 2   start with a number, make it smaller

two less than a number   x - 2   start with a number, make it smaller

four more than a number   x + 4 or 4 + x   start with a number, add 4

the difference between 6 and a number   6 - x   subtract, starting with 6

subtract four from a number   x - 4   starting with a number, subtract 4

the sum of 4 and a number   x + 4 or 4 + x   add to get a sum

the opposite of the sum of a number and three  -(x+3) or -(3+x)   take sum, take opposite of sum

     It may be convenient for the reader to use the technique the author used in writing the reason column.

Think of the word expression as a task to be completed by a computation.

Determine the operations needed.

Think how one would start and, then

think of what would need to be done to complete the task described.

     Notice the uses of the preposition "of" in the last example. Prepositional phrases break up the expressions into pieces. Those beginning with "of" are discussed on the last page.

     Below, notice how the prepositional phrases, in blue, which begin with by, between, from, and than, make it easier to examine the phrase and determine where to start and what to do.

words algebraic symbols reason

two decreased by a number   2 - x   start with 2, make it smaller

a number decreased by two   x - 2   start with a number, make it smaller

two less than a number   x - 2   start with a number, make it smaller

four more than a number   x + 4 or 4 + x   start with a number, add 4

the difference between 6 and a number   6 - x   subtract, starting with 6

subtract four from a number   x - 4   starting with a number, subtract 4

the sum of 4 and a number   x + 4 or 4 + x   add to get a sum

the opposite of the sum of a number and three  -(x+3) or -(3+x)   take sum, take opposite of sum

     Additional information is found in a prepositional phrases. The preposition "from" is almost always a tip-off that subtraction is required. Often, the desired mathematical expression starts with what is in the "from" prepositional phrase.

Multiplication & Division

     Here are words associated with multiplication or division. The context is the key to the symbols needed, not the word itself. Remember, the word AND is a signal that there are two or more ideas, not that addition is required. Remember, to use prepositional and other phrases to break up the expression in order to determine what is required. Examples follow the list.

divide

quotient

halve, half

third

product

double

triple

square

words algebraic symbols reason

the product of two and a number   2x or 2(x)   multiply

a third of the sum of a number and two   (x + 2)/3   start with the sum, take a third of it

double the difference between 6 and a number   2(6-x)   start with the difference, multiply by 2

triple the sum of four, five, and a number   3(4 + 5 + x)   start with the sum, multiply by 3

the value of 4 nickels   4(5) or 5+5+5+5, etc.   four fives or five four times

the value of some nickels   5x   multiply 5 times a number

     The last example may be HARD. Use the table below to examine the reason more clearly. The value of x nickels is 5x. The value of some nickels is 5x.

number of nickels value of nickels

1   5(1) = 5

2   5(2) = 10

3   5(3) = 15

4   5(4) = 20

x   5(x) = 5x

Exercises

I. Write the expression/equation then check the answer.

Note: Your browser may print the answers with spaces that are not ment to be there.

1.) the square of five    

Answers

5²

2.) the square of five subtracted from a number    

Answers

x - 5²

3.) the opposite of a number    

Answers

-x

4.) six less than half of a number    

Answers

x/2 - 6 or .5x - 6  

5.) the product of a number and four    

Answers

4(x) or 4x or (4)(x)

6.) a number decreased by five

Answers

x - 5  

7.) five less than a number  

Answers

x - 5  

8.) Ten is five less than a number.  

Answers

10 = x - 5  

9.) A number is less than five.

Answers

x < 5  

       See additional info on these.

10.) one less than a number  

Answers

x - 1

11.) one less than a double a number    

Answers

2x - 1

12.) the opposite of the sum of a number and 12    

Answers

-(x + 12)

13.) triple a number    

Answers

3x

14.) twice the sum of a number and 1    

Answers

2(x + 1)

III. Complete the table. Check each answer.

words algebraic symbols

1.  

Answers

one less than a number

 x - 1

2. the sum of six and double a number  

Answers

6 + 2x or 2x + 6

3. double the sum of six and a number    

Answers

2(x + 6)

4. the sum of three consecutive integers  

Answers

first + second + third

(x) + (x + 1) + (x + 2)  

5. the product of 2 consecutive odd integers    

Answers

(first)(second)  

(x)(x + 2)  

IV. Complete the table. Check each answer.

words algebraic symbols

1. one more than a number  

Answers

x + 1 or 1 + x

2.  

Answers

one less than a number or a number decreased by one

 x - 1

3. the product of a number and 3 and a different number  

Answers

3xy

4.  

Answers

double a number or the product of a number and two

 2x

5.  

Answers

the square of a number or a number multiplied by itself

 x2

Answer:

\(y=-4x+5\)

Step-by-step explanation:

The equation of a line with slope \(m\) and \(y\)-intercept \((0,b)\) is \(y=mx+b\).

1.1. Accumulated depreciation:

The accumulated depreciation for the machine bought on 1 July 2013 would be R150,000 as of 31 March 2016.

1.2. Machine realization:

The machine bought in 2013 was sold for R120,000 on 30 June 2015, resulting in a profit/loss on the sale of R10,000.

1.1. Accumulated Depreciation:

To calculate the accumulated depreciation, we need to determine the annual depreciation expense for each machine and then accumulate it over the years.

Machine bought on 1 July 2013:

Cost: R250,000

Depreciation rate: 20% per annum on cost

Depreciation expense for the year ended 31 March 2014: 20% of R250,000 = R50,000

Depreciation expense for the year ended 31 March 2015: 20% of R250,000 = R50,000

Depreciation expense for the year ended 31 March 2016: 20% of R250,000 = R50,000

Accumulated depreciation for the machine bought on 1 July 2013:

As of 31 March 2014: R50,000

As of 31 March 2015: R100,000

As of 31 March 2016: R150,000

1.2. Machine Realisation:

To record the sale of the machine bought in 2013, we need to adjust the machine's value and the accumulated depreciation.

Machine's original cost: R250,000

Accumulated depreciation as of 30 June 2015: R100,000

Net book value as of 30 June 2015:  

R250,000 - R100,000 = R150,000.

On 30 June 2015, the machine was sold for R120,000.

Realisation amount: R120,000

To record the sale:

Debit Cash: R120,000

Debit Accumulated Depreciation: R100,000

Credit Machine: R250,000

Credit Machine Realisation: R120,000

Credit Profit/Loss on Sale of Machine: R10,000 (difference between net book value and realisation amount).

These entries will reflect the appropriate balances in the ledger accounts and properly close off the accounts for the years ended 31 March 2016.

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Answer: D

Step-by-step explanation:

Hope it helps

The weight of the 0.8-meter piece is 19.6 kilograms.

We can use the ratio of length to weight to determine the weight of the 0.8-meter piece.

Let's call the weight of the 2.8-meter piece "W₁" and the weight of the 0.8-meter piece "W₂". Then we have:

W₁/2.8m = 24.5kg/1m

Solving for W₁, we get:

W₁ = (24.5kg/1m) x 2.8m = 68.6kg

Now we can use the same ratio to find W₂:

W₂/0.8m = 24.5kg/1m

Solving for W₂, we get:

W₂ = (24.5kg/1m) x 0.8m = 19.6kg

Therefore, the weight of the 0.8-meter piece is 19.6 kilograms.

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The Equivalent ratio to 8/2 is 4/1, 24/6, 32/8 and 48/12.

We have,

Cylinder and prism.

Here we have to find the equivalent ratio to 8/2.

First simplifying the given ratio as

8/2

= (4 x 2)/2

= 4 /1

Now, some more ratios equivalent to 8/2.

1. 8/2 x 3/3 = 24/ 6

2. 8/2 x 4/4 = 32/8

3. 8/2 x 6/6= 48/12

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Answer:21,849.24

Step-by-step explanation:

Answer:

9

Step-by-step explanation:

f(x) = x•x

f(-3) = -3*-3

f(-3)= 9

Answer:

56 minutes

Step-by-step explanation:

We can use proportions to solve.

6 miles            8 miles

------------     = ------------------

42 minutes        x minutes

Using cross products.

6x = 42 *8

Divide each side by 6

x = 42/6 * 8

x = 7*8

x = 56

To simplify the expression \(\displaystyle\sf 5x-3(-5y+6x)+4y\), we can follow the order of operations, which involves simplifying the expressions within parentheses, applying the distributive property, and combining like terms.

Using \(\displaystyle\sf \) tags for formatting, the expression becomes:

\(\displaystyle\sf 5x-3(-5y+6x)+4y\)

First, we simplify the expression within the parentheses:

\(\displaystyle\sf 5x-3(-5y+6x)+4y\)

\(\displaystyle\sf 5x+15y-18x+4y\)

Next, we can combine the like terms:

\(\displaystyle\sf 5x-18x+15y+4y\)

\(\displaystyle\sf -13x+19y\)

Therefore, the simplified expression is \(\displaystyle\sf -13x+19y\).

\(\huge{\mathfrak{\colorbox{black}{\textcolor{lime}{I\:hope\:this\:helps\:!\:\:}}}}\)

♥️ \(\large{\underline{\textcolor{red}{\mathcal{SUMIT\:\:ROY\:\:(:\:\:}}}}\)

Answer: log(2) 512= 9

Step-by-step explanation:

A P E X ;)

In logarithmic form, the expression \(2^9 = 512\) can be rewritten as follows:

\(log_{(2)}\ 512 = 9\)

As we know that the logarithmic form allows us to express the exponent as the logarithm of the base to a certain value, which in this case is 9.

To understand this, let's break down the expression.

The base of the logarithm is 2, and the result of raising 2 to the power of 9 is 512.

By taking the logarithm of 512 to the base 2, we get the exponent, which is 9.

So, in logarithmic form can be written as:

\(log_{(2)}\ 512 = 9\)

Therefore, in logarithmic form, the expression \(2^9 = 512\) can be rewritten as follows:

\(log_{(2)}\ 512 = 9\)

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Answer:

\(\frac{7}{2\sqrt{2} }\)

Step-by-step explanation:

substitute 2 in for x. the square root of 2 is already in radical form

Answer: 93

Step-by-step explanation:

To solve this problem, we need to use the order of operations, or PEMDAS.

Parenthesis

Exponent

Multiply

Divide

Add

Subtract

For multiply/divide and add/subtract, you don't necessarily go in that specific order. You go from left to right, depending on what comes first. For example, if division comes before multiply, you would divide first, then multiply. Same goes for add/subtract.

9²+3×(9-5)²/4                          [solve parenthesis]

9²+3×(4)²/4                              [solve exponent]

81+3×16/4                                [solve multiply or divide, whichever comes first]

81+12                                        [solve add or subtract, whichever comes first]

93

Now, we know that 93 is our final answer.

Answer:

1885.714

Step-by-step explanation:

I = ptr

5280 = 7×0.4×p

p = 5280/2.8

p = 1885.7

Answer:

The width is 410ft and the length is 780ft

Answer:

X= 110/7 or 15.7 or rounded to 16

Answer:

x=15.714285

Step-by-step explanation:

-0.35x=-3-2.5

-0.35x=-5.5

-0.35/-0.35= -5.5/0.35

X=-5.5/-0.35

X=15.714285

Answer:

P ≈ $12,654.89

Step-by-step explanation:

To calculate the amount of money that should be deposited today, we can use the formula for compound interest:

A = P(1 + r/n)^(nt)

Where:

A = the future value of the investment ($15,000 in this case)

P = the principal amount (the amount to be deposited today)

r = the annual interest rate (4.5% or 0.045 as a decimal)

n = the number of times the interest is compounded per year (monthly compounding, so n = 12)

t = the number of years (4 years in this case)

Substituting the given values into the formula, we have:

$15,000 = P(1 + 0.045/12)^(12*4)

Simplifying the equation:

$15,000 = P(1.00375)^(48)

To solve for P, we divide both sides of the equation by (1.00375)^(48):

P = $15,000 / (1.00375)^(48)

Using a calculator, we find:

P ≈ $12,654.89

Therefore, approximately $12,654.89 should be deposited today in order to accumulate to $15,000 in 4 years with a 4.5% annual interest rate compounded monthly.

The solution to the inequality f(x²-2) < f(7x-8) over D₁ = (-∞, 2) is:

-∞ < x < 1 or 1 < x < 6 or 6 < x < 2

Given the inequality,

    f(x²-2) < f(7x-8) over D₁ = (-∞, 2)

We need to find the values of x that satisfy this inequality.

Since we know that f is increasing over its domain, we can compare the values inside the function to determine the values of x that satisfy the inequality.

First, we can find the values of x that make the expressions inside the function equal:

x² - 2 = 7x - 8

Simplifying, we get:

x² - 7x + 6 = 0

Factoring, we get:

(x - 6)(x - 1) = 0

So the values of x that make the expressions inside the function equal are x = 6 and x = 1.

We can use these values to divide the domain (-∞, 2) into three intervals:

-∞ < x < 1, 1 < x < 6, and 6 < x < 2.

We can choose a test point in each interval and evaluate

f(x² - 2) and f(7x - 8) at that point. If f(x² - 2) < f(7x - 8) for that test point, then the inequality holds for that interval. Otherwise, it does not.

Let's choose -1, 3, and 7 as our test points.

When x = -1, we have:

f((-1)² - 2) = f(-1) < f(7(-1) - 8) = f(-15)

Since f is increasing, we know that f(-1) < f(-15), so the inequality holds for -∞ < x < 1.

When x = 3, we have:

f((3)² - 2) = f(7) < f(7(3) - 8) = f(13)

Since f is increasing, we know that f(7) < f(13), so the inequality holds for 1 < x < 6.

When x = 7, we have:

f((7)² - 2) = f(47) < f(7(7) - 8) = f(41)

Since f is increasing, we know that f(47) < f(41), so the inequality holds for 6 < x < 2.

Therefore, the solution to the inequality f(x²-2) < f(7x-8) over D₁ = (-∞, 2) is:

-∞ < x < 1 or 1 < x < 6 or 6 < x < 2

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Answer:

13.3

Step-by-step explanation:

Answer:

your wrong

Step-by-step explanation:

Answer:

8

Step-by-step explanation:

12-10=2

10-m=2

m=8

Answer: Volume = 40 cm3

Step-by-step explanation:

Answer:

-22

Step-by-step explanation:

\(1. \: 5 {}^{2} - (50 - 3) \\ 2. \: 5 {}^{2} - 47 \\ 3. \: 25 - 47 \\ 4. \: - 22\)

Given :

⠀

To Find :

⠀

Solution :

We know that,

\(\qquad { \pmb{ \bf{Length \times Width = Area_{(rectangle)}}}}\:\)

⠀

Now substituting the values :

\(\qquad {\dashrightarrow{ \sf{214 \: in\times 234 \: in = Area_{(rectangle)}}}}\:\)

\(\qquad {\dashrightarrow \: { { \sf{50,076 \: {in}^{2} = Area_{(rectangle)}}}}\:}

\)

⠀

Hence, The area of the rectangle is 50,076 in² .

The area of a rectangle with a length of 2 1/4 inches and a width of 2 3/4 inches is 6 3/16 in². Option C is correct.

Area of a rectangle is the product of the length of the rectangle and the width of the rectangle. It can be given as,

\(A=a\times b\)

Here, (a)is the length of the rectangle and (b) is the width of the rectangle.

The rectangle given in the problem has,

Thus, the area of the rectangle is,

\(A=2\dfrac{1}{4}\times2\dfrac{3}{4}\\A=\dfrac{8+1}{4}\times\dfrac{8+3}{4}\\A=\dfrac{9}{4}\times\dfrac{11}{4}\\A=\dfrac{99}{16}\\\A=\dfrac{96+3}{16}\\A=6\dfrac{3}{16}\)

Thus, the area of a rectangle with a length of 2 1/4 inches and a width of 2 3/4 inches is 6 3/16 in². Option C is correct.

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Using a geometric sequence, it is found that the maximum height reached after the 5th bounce is of 43.48 feet.

A geometric sequence is a sequence in which the result of the division of consecutive terms is always the same, called common ratio q.

The nth term of a geometric sequence is given by:

\(a_n = a_1q^{n-1}\)

In which \(a_1\) is the first term.

In this problem, the ball is dropped from a height of 101.00 inches onto a hard flat floor, and after each bounce, the height is 19% less, that is, 81% of the previous height, hence the first term and the common ratio are given by:

\(a_1 = 101, q = 0.81\)

Then, the height of the nth bounce is given by:

\(a_n = 101(0.81)^{n-1}\)

And the height of the 5th bounce is:

\(a_5 = 101(0.81)^{5-1} = 43.48\)

The maximum height reached after the 5th bounce is of 43.48 feet.

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Given:

Initial amount in the account = $1,000

Interest rate = 3.25%

Since the type of interest is simple, 3.25 % of 1000 will be added to that account every year with no increase in the incrementing amount

Amount increased in 18 years:

Amount increased every year = 3.25 * 1000 / 100  = $32.5

Amount increased in 18 years = amount gained in 1 year * 18

Amount increased in 18 years = 32.5 * 18 = $585

Total amount in John's bank account at the age of 18:

Initial amount + amount increased in 18 years

1000 + 585

$1,585

Answer:

Mandatory, prohibition, warning, danger, fire, emergency information, and limitation are the seven types of safety signage.

Step-by-step explanation: