Solve the following equation for x

5x-30y=-35.

Answer:

  |-3| = 3 and -|-4| = -4

Step-by-step explanation:

The absolute value function changes the sign to positive, if it isn't already. The usual rules of arithmetic and logic apply to these statements.

|-3| = 3 (true) and -|-4| = -4 (true)   ⇒ this statement is true

|-3| = 3 (true) and |-4| = -4 (false)   ⇒ this statement is false

-|3| = 3 (false) and |4| = 4 (true)   ⇒ this statement is false

|-3| = 3 (true) and -|4| = 4 (false)   ⇒ this statement is false

__

Additional comment

A compound "and" statement is only true if all of the parts of it are true.

Answer: 320 mL

Step-by-step explanation:

Given information

The bottle is currently 3/8 full

Current Volume = 240 mL

Concept:

Imagine the water in the bottle being separated into 8 parts

Currently, 3 parts are being occupied

Determine the volume for 1 part

3 parts = 240 mL

1 part = 240 / 3 = 80 mL

Determine the volume for half-full (4 parts)

1 part = 80 mL

4 parts = 80 × 4 = [tex]\Large\boxed{320~mL}[/tex]

Hope this helps!! :)

Please let me know if you have any questions

Given the following:

The economic order quantity is 24.

Eoq = √2 * demand * ordering cost / holding cost)

= √(2 * 1050 * 15 / 55)

EOQ = 24

The  Average inventory = EOQ / 2

= 24 / 2

= 12

Expected number of orders = demand / EOQ

= 1050 / 24

= 44

Annual holding cost (AHC) = (EOQ / 2) * Holding cost

= (24 / 2) * 55

= 660

Annual ordering cost (AOC) = (demand / EOQ) * ordering cost

= (1050 / 24) * 15

= 656

Thus,

Total cost of managing = AHC + AOC

= 660 + 656

TCM = 1,316

5. For holding cost of 60

EOQ= √(2 * demand * ordering cost / holding cost)

= √(2 * 1050 * 15 / 60)

= 23

Annual holding cost = (EOQ / 2) * holding cost

this gives us

= (23 / 2) * 60

= 690

Annual ordering cost = (demand / EOQ) * ordering cost

= (1050 / 23) * 15

= 685

Total cost of managing = AHC+ AOC

= 690 + 685

= 1375

Change in total annual cost = new - old

= 1375 - 1316

= 59

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Answer:

Step-by-step explanation:

answer: 47

add the heights then divide by 5

Answer:

IT HAS A SLOPE OF -2

Y INTERCEPT IS 0,0

Step-by-step explanation:

Answer:

  x = 3

Step-by-step explanation:

Maybe you want the value of x such that ...

  4(6^x) = 864

Dividing by 4 gives ...

  6^x = 216

You may know that 216 = 6^3. Using that, we can equate exponents:

  6^x = 6^3

  x = 3

Alternatively, we can use logarithms to find x. Taking logs gives ...

  x·log(6) = log(216)

  x = log(216)/log(6) = 3

Answer: 9

Step-by-step explanation: it is what it is

15-6=9 omg i never knew

The second number line best shows how to solve –4 – (–8). This can be obtained by finding the value of –4 – (–8) and checking which number line has the required arrow mark.

The value of  –4 – (–8) is obtained.

–4 – (–8) =  –4 + 8

–4 – (–8) = 4

The arrow should move from zero to - 4 and the second arrow should move from - 4 to 4.

From the question given the number lines,

The first arrow is moving from zero to - 4.

First condition is satisfied.

The second arrow is moving from - 4 to  8.

Second condition is not satisfied.

The first arrow is moving from zero to - 4.

First condition is satisfied.

The second arrow is moving from - 4 to  4.

Second condition is satisfied.

Hence the second number line best shows how to solve –4 – (–8).

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The magnitude of -1-5i is √26.

Given the complex expression; -1 - 5i

To find the magnitude, we use the formula;

| a+bi | = √[ a² + b² ]

| a+bi | = √[ a² + b² ]

| -1-5i | = √[ (-1)² + (-5)² ]

| -1-5i | = √[ 1 + 25 ]

| -1-5i | = √26

Therefore, the magnitude of -1-5i is √26.

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The value of k such that the graph of f and the graph of g only intersect one is equal to 2.

The value of k such that the graph of f and the graph of g only intersect one is equal to 2. According to the image attached below, functions f(x) and g(x) intersect at point (x, y) = (0, 4) for k = 2.

Herein we have a system formed by two nonlinear equations, a quadratic equation and a cubic equation. Given the constraint that both function must only intersect once, we have the following expression:

f(x) - x² - k · x = 4       (1)

g(x) - x² - k · x = x³ + 2 · k        (2)

x³ + 2 · k = 4

x³ + 2 · (k - 2) = 0

If f and g must intersect once, then the roots must of the form:

(x - r)³ = x³ + 2 · (k - 2)

x³ - 3 · r · x² + 3 · r² · x - r³ = x³ + 2 · (k - 2)

Then, the following conditions must be met: - 3 · r · x² = 0, 3 · r² · x = 0. If x may be any real number, then r must be zero and the value of k must be:

2 · (k - 2) = 0

k - 2 = 0

k = 2

Therefore, the value of k such that the graph of f and the graph of g only intersect one is equal to 2. According to the image attached below, functions f(x) and g(x) intersect at point (x, y) = (0, 4) for k = 2.

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6 decks of cards = 60 cards

---> 60/8 = 15/2 = 7.5

7.5 cards = 0.75 decks of cards

Therefore, each student will get 0.75 decks of cards.

Answer:

1

Step-by-step explanation:

Everyone gets 1 and there's 2 left over

Answer:

[tex]3x^{2} - 4xy + 2y { }^{2} - 1 \\ 3 \times ( - 3) { }^{2} - 4 \times ( - 3) \times 5 + 2 \times 5 {}^{2} - 1 \\ (3 \times 9) - ( - 60) + 50 - 1 \\ 27 + 60 + 50 - 1 \\ 165 [/tex]

Answer: 136

Substitute -3 for x and 5 for y.

[tex]3x^2 - 4xy + 2y^2 - 1[/tex]

[tex]3(-3)^2-4(-3)(5)+2(5)^2-1[/tex]

[tex]3(9)-4(-15)+2(25)-1[/tex]

[tex]27+60+50-1[/tex]

[tex]87+50-1[/tex]

[tex]137-1[/tex]

[tex]136[/tex]

hope this helped!

Answer:

e

Step-by-step explanation:

[tex]volume~of~cone=\frac{1}{3}volume ~of~cylinder\\volume~of~cylinder=3 \times volume ~of~cone=3 \times15=45 \\volume ~of~two~cylinders=2 \times45=90[/tex]

The company's inventory turnover ratio, based on the financial data, equals 10 times.

The inventory turnover ratio measures the number of times the company sells and replaces its inventory over a given period.

The formula for calculating the Inventory Turnover Ratio is the Cost of Goods Sold divided by the Average Inventory.

The Average Inventory is the mean of the beginning and ending inventories.

Sales revenue = $60,000

Cost of goods sold = $20,000

Beginning inventory = $1,600

Ending inventory = $2,400

Average inventory = $2,000 ($1,600 + $2,400)/2

Inventory turnover ratio = 10x ($20,000/$2,000)

Thus, the company sells and replaces its inventory over a period of 10 times.

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1. Given the functions f(x) = log₂(5x)  and g(x) = 5ˣ - 2 only statement c is true

2. The solution of 3y² + 5y > -2 is a. x < –1 or x is greater than negative two thirds

Statement a

Since f(x) = log₂(5x) which is a logarithm function is undefined for (-∞, 0) and defined for (0, +∞) and g(x) = 5ˣ - 2 which is an exponential function is defined for (-∞, +∞).

Also, since f(x) is decreasing on the interval (0, 1/5) while g(x) decreases on the interval (-∞, 0). So, they have do not have a common interval on (0, 1).

So, statement a. Both f(x) and g(x) decrease on the interval of (–∞, 1).

is false

Statement b

Since f(x) = log₂(5x) which is a logarithm function is defined for (0, +∞)  and g(x) = 5ˣ - 2 which is an exponential function is defined for (-∞, +∞).

So, the statement b Both f(x) and g(x) have the same domain of (0, ∞) is false

Statement c

Since f(x) = log₂(5x) which is a logarithm function has a range of (0, +∞).  and g(x) = 5ˣ - 2 which is an exponential function is has a range of (-2, +∞).

So, they have a common interval of (0, +∞).

So, the statement c. Both f(x) and g(x) have a common range on the interval (–2, ∞) is true

Statement d

To find the x-intercept of f(x), we equate f(x) to zero.

So, f(x) = log₂(5x)

0 = log₂(5x)

2⁰ = 5x

1 = 5x

x = 1/5

To find the x-intercept of g(x), we equate g(x) to zero.

g(x) = 5ˣ - 2

0 = 5ˣ - 2

2 = 5ˣ

x = ㏒₅2

Since the x-intercept of f(x) = 1/5 and the x- intercept of g(x) = ㏒₅2. So, they do not have a common x - intercept.

So, the statement d. Both f(x) and g(x) have the same x-intercept of (1, 0) is false.

So, only statement c is true

3y² + 5y > -2

3y² + 5y + 2 > 0

3y² + 3y + 2y + 2 > 0

3y(y + 1) + 2(y + 1) > 0

(3y + 2)(y + 1) > 0

So, the boundary values are at

(3y + 2)(y + 1) = 0

(3y + 2) = 0 or (y + 1) = 0

y = -2/3 or y = -1

So, we require (3y + 2)(y + 1) > 0

For y < -1 say -2, (3y + 2)(y + 1) = (3(-2) + 2)((-2) + 1)

= (-6 + 2)(-2 + 1)

= -4(-1)

= 4 > 0

For -1 < y < -2/3 say -1/3, (3y + 2)(y + 1) = (3(-1/3) + 2)((-1/3) + 1)

= (-1 + 2)(-1 +3)/2

= 1(-2/2)

= -1 < 0

For y > -2/3 say 0, (3y + 2)(y + 1) = (3(0) + 2)((0) + 1)

= (0 + 2)(0 + 1)

= 2(1)

= 2 > 0

So, for (3y + 2)(y + 1) > 0, y < -1 or y > -2/3

So, the solution of 3y² + 5y > -2 is a. x < –1 or x is greater than negative two thirds

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Answer:

1. D. Both f(x) and g(x) have the same x-intercept of (1, 0).

2. A. x < –1 or x is greater than negative two thirds

Step-by-step explanation:

I took the exam

Answer:

4.93

See the discussion below if the number is 4928.

Step-by-step explanation:

Comment

Do you really mean hundred? If you do, the nearest hundred is 0 when the actual value is 4.928

If you mean the nearest hundredth that means that 2 is the value of the hundredth now. Rounded, the answer is 4.93

If the given value is 4928, the hundred's place is the 9. Look at the 10s place (2). Since the 2 is less than 5, the 9 is not changed. 2 zeros are at the end of 49 since the hundreds place is rounded.

4900 is the answer

Hi there,

please see below for solution steps :

__________

Let us revise the rounding rules :

1) if the number you’re rounding to is followed by a number from 0 to 4, you drop that number

2) if the number you’re rounding to is followed by a number equal to 5 or more, you add 1 to that number

________

Now,

Nearest Hundredth => 3 digits after the decimal point

Thus,

4.928 => 4.93

Hope the Answer - and steps - made sense to you,

Happy Studying !!

Frozen Melody

The sum of the two rational equations is equal to (3 · n² + 5 · n - 10) / (3 · n³ - 6 · n²).

In this question we must use algebra definitions and theorems to simplify the addition of two rational equations into a single rational equation. Now we proceed to show the procedure of solution in detail:

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2 angles that have the same angle measure are congruent.

Given that the two angles have the same angle measure.

Congruence is a term reserved for geometry. Two metrics are congruent when you can perfectly map to each other by mirroring, rotating, and translating without distortion.

We know that two shapes or objects are congruent if they have the same shape and the same size. For two angles to be congruent, they must coincide when they are overlapped. This is only possible if the two angles are equal.

Therefore, if two angles have the same measure, then they are said to be congruent.

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The formula that can be used to illustrate consecutive even numbers is a + 2.

It should be noted that the information is incomplete. Therefore, an overview will be given.

It should be noted that consecutive even numbers we the numbers that follow each other and have a difference of 2.

Examples of such numbers include 2, 4, 6,8, 10, etc.

In this case, the formula that can be used to illustrate consecutive even numbers is a + 2.

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The first five terms of the sequence representing each bottle scheme for Mike is illustrated below.

From the information given, Mike and Tim are looking to earn a little extra money and the beach committee offers them the opportunity of picking up plastic bottles, paying them $0.20 per bottle.

For Mike, the price for the first bottle is $0.10.

The price for the second bottle will be:

= 0.10 × 1.02 = 0.102

The price for the third bottle will be:

= 0.10 × 1.02² = 0.10404

The price for the forth bottle will be:

= 0.10 × 1.02³ = 0.10612

The price for the fifth bottle will be:

= 0.10 × 1.02⁴ = 0.10824

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Complete question:

Mike and Tim are looking to earn a little extra money. The beach committee offers them the opportunity of picking up plastic bottles, paying them $0.20 per bottle. Mike realizes as they pick up it will get harder and harder to find more bottles. So as an incentive to keep looking he suggests a different form of payment. He suggests $0.10 for the first bottle and increase the pay by 2% for each bottle after that. Tim thinks Mike is crazy to propose an increase of just 2% per piece. He plans to ask for a one cent increase for every piece, starting at 15 cents for the first bottle. Write the first five terms of the. sequence representing each bottle scheme for Mike.

(i) The expanded form of (1 / 2 - 2 · x)⁵ in ascending form is 1 / 32 - (5 / 8) · x + 5 · x² - 20 · x³ + 40 · x⁴ - 32 · x⁵.

(ii) The coefficient of x³ from (1 + a · x + 3 · x²) · (1 / 2 - 2 · x)⁵ is - 265 / 8.

In this problem we must apply the concept of Pascal's triangle to expand the power of a binomial of the form (x + y)ⁿ and further algebra properties.

(i) First, we proceed to expand the power binomial (1 / 2 - 2 · x)⁵ in ascending order:

(1 / 2 - 2 · x)⁵ = (1 / 2)⁵ + 5 · (1 / 2)⁴ · (- 2 · x) + 10 · (1 / 2)³ · (- 2 · x)² + 10 · (1 / 2)² · (- 2 · x)³ + 5 · (1 / 2) · (- 2 · x)⁴ + (- 2 · x)⁵

( 1 / 2 - 2 · x)⁵ = 1 / 32 - (5 / 8) · x + 5 · x² - 20 · x³ + 40 · x⁴ - 32 · x⁵

(ii) Second, we proceed to expand the following product of polynomials by algebra properties:

(1 + a · x + 3 · x²) · (1 / 2 - 2 · x)⁵ = (1 + a · x + 3 · x²) · [1 / 32 - (5 / 8) · x + 5 · x² - 20 · x³ + 40 · x⁴ - 32 · x⁵]

(1 + a · x + 3 · x²) · (1 / 2 - 2 · x)⁵ = 1 / 32 + (a / 32 - 5 / 8) · x + (- 5 · a / 8 + 163 / 32) · x² + (- 175 / 8 + 5 · a) · x³ + (65 - 20 · a) · x⁴ + (- 92 + 40 · a) · x⁵ + (120 - 32 · a) · x⁶ - 96 · x⁷

In accordance with the statement, we find that:

- 5 · a / 8 + 163 / 32 = 13 / 2

- 5 · a / 8 = 45 / 32

a = - 9 / 4

Thus, the coefficient of x³ is:

C = - 175 / 8 + 5 · (- 9 / 4)

C = - 265 / 8

The coefficient of x³ from (1 + a · x + 3 · x²) · (1 / 2 - 2 · x)⁵ is - 265 / 8.

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The function that has a minimum and is transformed to the right and down from the parent function is; g(x) = 8(x² - 3²) - 5

By inspection, only Functions B) and D) have a minimum.

For the first function in option B, we have:

g(x) = 4(x²- 6x + 9) + 1

This can be simplified to;

g(x) = 4( x - 3 )² + 1

Thus, we can say that this first function is transformed to the right and up from the parent function.

For the second function in option D, we have;

g(x) = 8(x² - 6x + 9 ) - 5

The function can be simplified to get;

g(x) = 8( x - 3 )² - 5

Thus, we can say that it is transformed to the right and down.

The function that has a minimum and is transformed to the right and down from the parent function is D

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Answer: option D

Step-by-step explanation:

Answer:

y = (constant variable x time) + independent variable

y = 60x + 30    ← Health Club B

y = 27x + 50    ← Health Club A

Step-by-step explanation:

I can't see the question at the bottom due to the covering but if you comment I can edit this and give a proper answer. Otherwise this is all I can give you.

For a standard normal distribution, the probability of the 2 - scores P(-1.64 < z < 0.2) is 0.52876

We want to find the p-value between 2 z-scores expressed as;

P(-1.64 < z < 0.2)

To solve this, we will solve it as;

P(-1.64 < z < 0.2) = 1 - [P(z < -1.64)  + P(z > 0.2)]

From normal distribution table, we have that;

P(x < -1.64) = 0.050503

P(x > 0.2) = 0.42074

Thus;

P(-1.64 < z < 0.2) = 1 - (0.050503 + 0.42074)

P(-1.64 < z < 0.2) = 0.52876

Thus, For a standard normal distribution, the probability of the 2 - scores P(-1.64 < z < 0.2) is 0.52876

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Answer:

1) 13m, 2) 10ft, 3) 11.4ft, 4) 12.7m

Step-by-step explanation:

Since all of these triangles are right triangles, we can use the Pythagorean Theorem: a^2+b^2 = c^2

1) 5^2+12^2 = c^2 = 25+144 = c^2 = 169 = c^2 = 13 = c

2) 6^2+8^2 = c^2 = 36+64 = c^2 = 100 = c^2 = 10 = c

3) 5^2+10.3^2 = c^2 = 25+106.09 = c^2 = 131.09 = c^2 = 11.4494541 = 11.4 = c

4) 8.6^2+9.4^2 = c^2 = 73.96+88.36 = c^2 = 162.32 = c^2 = 12.7404866 = 12.7 = c

The length of the diagonal that he cut to the nearest whole inch is 36 inches

A red dotted line crosses the rectangle from the upper left corner to the lower right corner to form a triangle.

Length of the diagonal of a rectangle;

Hypotenuse² = adjacent² + opposite²

= 18² + 36²

= 324 + 1296

hyp² = 1620

hyp = √1620

hyp = 35.4964786985976

Approximately,

hypotenuse = 36 inches

Therefore, the length of the diagonal that he cut to the nearest whole inch is 36 inches.

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The difference between Alok and Prabir share exists 8000.

Given: Invested by Alok = Rs 90, 000

Invested by Prabir = Rs 1,20000

The time period of Alok = 3 months

The time period of Prabir = 2 years

They earn a profit = Rs 96,000.

Profit exists directly proportional to the product of the amount invested and the time period of investment.

8000 = 90000 [tex]*[/tex] 24/120000 [tex]*[/tex] 21

= 5/7

5x + 7x = 96000

x = 8000

first = 40000

second = 48000

so the difference exists at 8000.

The difference between Alok and Prabir share exists 8000.

Therefore, the correct answer is 8000.

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The equation of parabola is a [tex]\frac{-(x+ 3)^{2}}{64} + \frac{ (y-1)^{2}}{4} = 1[/tex] and the coordinates of center is (0,0) And asymptote is become 16.

According to the statement

we have to find the centers, foci, vertices and asymptotes of the hyperbola from the given equations.

So, For this purpose,

Hyperbola is a a plane curve generated by a point so moving that the difference of the distances from two fixed points is a constant.

So, The given equation is:

16y²-x² - 6x - 32y = 57

So, rearrange it then

-x² - 6x - 32y + 16y² = 57

-(x² + 6x) + (16y² - 32y ) = 57

-(x² + 6x) + 16(y² - 2y ) = 57

-(x² + 6x + 9) + 16(y² - 2y +1 ) = 57 +1(16) + 9(-1)

-(x+ 3)²  + 16 (y-1)²  = 64

-(x+ 3)²  + 16 (y-1)²  = 64

divide whole equation by 64 then

[tex]\frac{-(x+ 3)^{2}}{64} + \frac{16 (y-1)^{2}}{64} =\frac{64}{64}[/tex]

then equation become

[tex]\frac{-(x+ 3)^{2}}{64} + \frac{ (y-1)^{2}}{4} = 1[/tex]

This become the equation of hyperbola.

So,

here coordinates of center is (0,0)

And asymptote is become 16.

So, The equation of parabola is a [tex]\frac{-(x+ 3)^{2}}{64} + \frac{ (y-1)^{2}}{4} = 1[/tex] and the coordinates of center is (0,0) And asymptote is become 16.

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Answer:

Step-by-step explanation:

[tex]2x > 30+\frac{5}{4x} \\2x-\frac{5}{4x} > 30\\\frac{8x^2-5}{4x} > 30\\case~1\\if~x > 0\\8x^2-5 > 120x\\8x^2-120x > 5\\x^2-15x > \frac{5}{8} \\adding~(-\frac{15}{2} )^2~to~both~sides\\(x-\frac{15}{2} )^2 > \frac{5}{8}+\frac{225}{4} \\(x-\frac{15}{2} )^2 > \frac{455}{8} \\x-\frac{15}{2} < -\sqrt{\frac{455}{8} } \\x < \frac{15}{2}-\sqrt{\frac{455}{8} } \\or~x < 0\\rejected~as~x > 0[/tex]

[tex]x-\frac{15}{2} > \sqrt{\frac{455}{8} } \\x > \frac{15}{2} +\sqrt{\frac{455}{8} }[/tex]

case~2

[tex]if~x < 0\\8x^2-5 < 120x\\8x^2-120x < 5\\x^2-15x < \frac{5}{8} \\adding~(-\frac{15}{2} )^2\\(x-\frac{15}{2} )^2 < \frac{5}{8} +(-\frac{15}{2} )^2\\|x-\frac{15}{2} | < \frac{5+450}{8} \\-\sqrt{\frac{455}{8} } < x-\frac{15}{2} < \sqrt{\frac{455}{8} } \\\frac{15}{2} -\sqrt{\frac{455}{8} } < x < \frac{15}{2} +\sqrt{\frac{455}{8} } \\but~x < 0\\7.5-\sqrt{\frac{455}{8} } < x < 0[/tex]

The count of the denominations of the bills of $1, $5, and $10, are 15, 10, and 5, respectively.

In the question, we are given that there are 3 denominations of bills in a wallet: $1, $5, and $10. There are 5 fewer $5-bills than $1-bills. There are half as many $10-bills as $5-bills.

We are asked to find the count of each denomination, given there was altogether $115 in the bag.

We assume the number of $1-bills in the bag to be x.

Total amount in x bills of $1 =  x * $1 = $x.

Given that there are 5 fewer $5-bills than $1-bills in the bag, number of $5-bills in the bag = x - 5.

Total amount in (x - 5) bills of $5 = (x - 5) * $5 = $5(x - 5).

Given that there are half as many $10-bills as $5-bills in the bag, number of $10-bills in the bag = (x - 5)/2.

Total amount in (x - 5)/2 bills of $10 = (x - 5)/2 * $10 = $5(x - 5).

Thus, the total amount in the bag = $x + $5(x - 5) + $5(x - 5).

But, the total amount in the bag = $115.

Thus, we get an equation:

$x + $5(x - 5) + $5(x - 5) = $115,

or, x + 5x - 25 + 5x - 25 = 115,

or, 11x = 115 + 50,

or, 11x = 165,

or, x = 165/11,

or, x = 15.

Thus, number of $1-bills = x = 15.

The number of $5-bills = x - 5 = 15 - 5 = 10.

The number of $10-bills = (x - 5)/2 = (15 - 5)/2 = 10/2 = 5.

Thus, the count of the denominations of the bills of $1, $5, and $10, are 15, 10, and 5, respectively.

The provided question is incomplete. The complete question is:

There are 3 denominations of bills in a wallet: $1, $5, and $10. There are 5 fewer $5-bills than $1-bills. There are half as many $10-bills as $5-bills. If there is $115 altogether, find the number of each type of bill in the wallet."

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