Transactions on Furnell's credit card are shown for the month of June. The interest rate is 1.4% per month.

June 1 Balance $352.12
June 4 Sears $331.89
June 8 eBay $81.58
June 15 Outback $30
June 18 Payment $200




Find the average daily balance $
184.89

Find the interest for the month $

Find the balance for the following month $

Answer:

The answer is A.

Step-by-step explanation:

For this graph, you are using the linear equation formula: y = mx + b

Answer:

N = M + 10/3P

Step-by-step explanation:

Answer:

3π

Step-by-step explanation:

Given

Radius=9

Angle=60°

Since PQ is an arc

Perimeter of an Arc =(θ/360)×2πr

60/360×2π(9)

1/6×18π

3π

Answer:

its 112.7

Step-by-step explanation:

x+2x=3x. 3x = 338

divide 3x and 338 by 3 three is cancelled and x equals to 112.7

[tex]\sf x=13.[/tex]

Assuming that the canvas has a standard rectangle shape, the area should be given by: [tex]\sf A= bh[/tex], where "b" is the base length of the rectangle, and "h" its heignt.

We're told that the canvas is "x" inches wide, and that the length of it is "twice as long as the width", thus it is "2x". Now, let's take width as the base, and length as the height.

Since the formula for area is just a product between width and length, and those values are "x" and "2x", respectively, the formula for the area of this canvas is:

[tex]\sf (x)(2x)=338(in^{2} )[/tex]

a) Solve the product.

[tex]\sf (x)(2x)=338(in^{2} )\\ \\\sf 2x^{2} =338(in^{2} )[/tex]

b) Divide by "2" on both sides of the equation.

[tex]\sf \dfrac{2x^{2}}{2} =\dfrac{338(in^{2} )}{2} \\ \\x^{2} =169[/tex]

c) Take the square root of both sides.

[tex]\sf \sqrt{x^{2}} =\sqrt{169} \\ \\x=13[/tex]

[tex]\sf x=13.[/tex]

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The sum of both numbers is 107/12

One number is 6 3/4

The other smaller number is 2 1/6

The sum of both numbers can be calculated as follows

= 6 3/4 + 2 1/6

= 27/4 + 13/6

= 162 + 52/24

= 214/24

= 107/12

Hence the sum of both numbers is 107/12

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Step-by-step explanation:

I think you mean the sun's mass is 1.99 × 10³⁰ kg.

70% of that is hydrogen :

1.99 × 10³⁰ × 0.7 = 1.393 × 10³⁰ kg

a.

13% of that is in the core :

1.393 × 10³⁰ × 0.13 = 0.18109 × 10³⁰ = 1.8109 × 10²⁹ kg

b.

600 000 000 000 kg = 6.0 × 10¹¹ kg/s

so, 1.8109 × 10²⁹ kg will last

1.8109 × 10²⁹ / (6.0 × 10¹¹) = 0.301816666... × 10¹⁸ =

= 3.01816666... × 10¹⁷ seconds

60 seconds per minute.

60 minutes per hour.

24 hours per day.

365 days per year.

1 000 000 000 years.

that is

31,536,000,000,000,000 seconds in 1 billion years =

= 3.1536 × 10¹⁶ seconds in 1 billion years

so, this will go on for

3.018166666... × 10¹⁷ / 3.1536 × 10¹⁶ billion years =

= 0.957054372... × 10¹ = 9.57054372... billion years.

c.

the sun has already "spent" 4.6 billion years, so, that leaves about

9.6 - 4.6 = 5 billion years

According to the information, the total mass of hydrogen available for nuclear fusion in the Sun's core is 10^29 kg, the duration of the oroginal supply of hydrogen of the sun is 3.02 × 10^17 seconds, and the left time of the sun is 4.98 billion years.

a. The total mass of hydrogen available for nuclear fusion in the Sun's core is:

0.7 × 1.99 × 10^30 kg × 0.13 = 1.81 × 10^29 kg

How long the sun's original suply of hydrogen can last?

t = (1.81 × 10^29 kg) / (6 × 10^11 kg/s) = 3.02 × 10^17 seconds

Converting to billions of years:

t = (3.02 × 10^17 s) / (3.15 × 10^16 s/billion years) = 9.58 billion years

c. The Sun has already burned for 4.6 billion years, so the amount of time it has left before it runs out of fuel is:

9.58 billion years - 4.6 billion years = 4.98 billion years

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Answer:

Step-by-step explanation:

The given total area including the frame is y = (8+2x)(10+2x).

As per the question the thickness of the frame is x.

The area of a rectangle is given by A=L*B

For a picture with frame it will be A=(L+2t)*(B+2t)

1. Comparing the above equation with the given equation it can be concluded that the Length is 10 and the width is 8.

2. The given equation is 100=(8+2x)(10+2x)

Upon solving it simplifies to [tex]x^2+9x-5=0[/tex]

x=0.525.

Thus the thickness x=0.525.

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(a) The domain of f of g is {1, 8}.

(b) The range of f of g is {0, 1}.

The domain of f of g consists of all the inputs in the domain of g that are also in the domain of f.

(a) Domain of f of g:

The inputs in the domain of g that are also in the domain of f are 1, and 8. Therefore, the domain of f of g is {1, 8}.

To find the range of f of g, we need to apply the function composition f(g(x)) to each input in the domain of f of g, and collect all the outputs.

(b) Range of f of g:

The range of f of g consists of all the outputs obtained by applying f(g(x)) to each input in the domain of f of g.

We have:

f(g(1)) = f(8) = 0

f(g(4)) = f(2) = 1

f(g(8)) = f(0) = 1

Therefore, the range of f of g is {0, 1}.

Thus, in set notation, the domain of f of g is {1, 8}, and the range of f of g is {0, 1}.

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The length of BC, to the nearest tenth of a centimeter, is calculated as: 12.6 centimeter.

Recall that two similar triangles will always have side lengths that are proportional in length. The triangles shown are similar to each other, therefore, the proportion below would be true:

AC/CE = BC/CD

Given the following:

AC = 7.2 units

CE = 2.4 units

CD = 4.2 units

BC = ?

Plug in the values:

7.2/2.4 = BC/4.2

Cross multiply:

BC = 7.2 * 4.2 / 2.4

BC = 12.6 cm

Thus, the length of side BC in the similar triangles shown is approximately 12.6 cm.

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Answer:

x+y=9

5x+8y=66

Step-by-step explanation:

This statement describes the relationship between symmetries and rotations

This statement describes a relation between two symmetry operations, denoted by [tex]s_i[/tex] and [tex]s_j[/tex], and a rotation operation denoted by [tex]r_{i-j}[/tex]. The notation [tex]"s_i \circ s_i"[/tex] and [tex]"s_j \circ s_i"[/tex] denotes the composition of these operations, which means applying one operation after the other.

The statement says that [tex]s_i[/tex] and [tex]s_j[/tex] are both symmetries, and that they have a specific relationship with each other: when [tex]s_i[/tex] is composed with itself, it gives the identity operation (denoted by 1), and when [tex]s_j[/tex] is composed with [tex]s_i[/tex], it also gives the identity operation.

Then the statement concludes that this relationship implies that [tex]s_j[/tex] can be obtained by first applying [tex]s_i[/tex] and then applying a rotation operation [tex]r_{i-j}[/tex], and that [tex]s_i[/tex] can be obtained by first applying a rotation operation [tex]r_{i-j}[/tex] and then applying [tex]s_j[/tex].

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Answer:

9 belongs to set N, natural, whole, integer, rational, and real number

Step-by-step explanation:

* Natural number: use for counting

* Whole number: includes natural number and zero

* Integer : are whole numbers and their opposite

*Rational number:come from fraction, and their results can be positive integer

*Real number: includes all numbers

The calculated output value for an input value of 5 is (c) 10

From the question, we have the following parameters that can be used in our computation:

Kerrin is making an input-output table for the rule "add 5."

This means that

Output = input + 5

In this case, we have

input = 5

substitute the known values in the above equation, so, we have the following representation

Output = 5 + 5

Evaluate

Output =10

Hence, the output is 10

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Answer:{y,x}={-4,-1}

Step-by-step explanation:

Answer:

Step-by-step explanation:

To determine which flight will have the most no-shows, we need to compare the probabilities of no-shows for each flight.

The probability of a no-show for the afternoon flight is 1/10, which means that on average, 1 out of every 10 passengers on the afternoon flight will not show up.

The probability of a no-show for the morning flight is 1/20, which means that on average, 1 out of every 20 passengers on the morning flight will not show up.

Since 1/10 is greater than 1/20, the afternoon flight will have a higher number of no-shows, on average. Therefore, the afternoon flight is more likely to have the most no-shows.

the rate of interest is 11.25%.we can use simple interest formula get answer as above

what is simple interest ?

Simple interest is a type of interest that is calculated as a percentage of the principal amount of a loan or investment, and is based only on the original amount of money that was borrowed or invested.

In the given question,

We can use the formula for simple interest to solve for the rate:

Simple Interest = (Principal x Rate x Time) / 100

Substituting the given values, we get:

$900 = (400 x Rate x 20) / 100

Multiplying both sides by 100 and dividing by 400 x 20, we get:

Rate = $900 / (400 x 20) = 0.1125 or 11.25%

Therefore, the rate of interest is 11.25%.

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The use of calculations to show that the height of the barrel of oil is 96.82cm is given below:

The volume is given as V = 42 gal

The radius is given as r = 18/2 = 9 in

The height is given as h= 96.83 cm

Using the formula:

V = πr²h

h = V/(πr²)

The volume is given as V = 42 gal

42 gal x 3.7854 l/gal = 158.987 l

158.987 l = 158,987 ml = 158,987 cm³

The radius is given as r = 18/2 = 9 in

9 in x 2.54 cm/in = 22.86 cm

The height is given as h = (158987 cm³) / π(22.86 cm)² ≈ 96.82 cm

depending on how you round, the more precise answer is 96.8285 ≈ 96.83 cm

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Answer: 15y2 - 19y + 6 = 0

Step-by-step explanation:

The first term is,  15y2  its coefficient is  15 .The middle term is,  -19y  its coefficient is  -19 .The last term, "the constant", is  +6 Step-1 : Multiply the coefficient of the first term by the constant   15 • 6 = 90 Step-2 : Find two factors of  90  whose sum equals the coefficient of the middle term, which is   -19 .

The integer which is closest to the radius of a circle with a circumference of 50.27 as required is; Choice A; 8.

It follows from the task content that the integer which is closest to the radius of the circle whose circumference is; 50.27 is to be determined.

Recall; Circumference, C = 2πr

r = C / 2π

r = 50.27 / 2π

r = 8.000718989

Ultimately, the integer which is closest to the radius is; Choice A; 8.

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The temperature at 1pm is -3°C, at 2pm it's -6°C, at 3pm it's -9°C, and at 6pm it's -18°C. The temperature t hours after noon is T = 0 - 3t

If the temperature is dropping at a rate of 3 degrees per hour, then the temperature t hours after noon can be represented by the equation T = 0 - 3t, where T is the temperature in degrees Celsius.

To find the temperature at a specific time after noon, we need to substitute the corresponding value of t into the equation and solve for T. For example:

At 1pm (t = 1), T = 0 - 3(1) = -3°C

At 2pm (t = 2), T = 0 - 3(2) = -6°C

At 3pm (t = 3), T = 0 - 3(3) = -9°C

At 6pm (t = 6), T = 0 - 3(6) = -18°C

To find the temperature t hours after noon, we can use the same equation: T = 0 - 3t. For example, if we want to find the temperature 4 hours after noon, we can substitute t = 4 into the equation: T = 0 - 3(4) = -12°C. So the temperature 4 hours after noon is -12°C.

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The result of adding -4 (row 1) to row 3 is determined as (0  - 9   2)|12.

The resultant of the row multiplication in the Matrice is calculated by applying the following method;

row 1 in the given matrices = [1  2  1] | -5

To multiply row by -4, we will multiply each entity by 4 as shown below;

= -4(1 2  1) | -5

= (-4  -8  - 4) |20

To add the result to 3;

(-4  -8  - 4)|20 + (4  - 1   6) | -8

= (0  - 9   2)|12

Thus, the result of the row multiplication is determined by multiplying each entry in row 1, by - 4.

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There was no error in the procedure as we can see below.

In algebra, a formula expresses a relationship between two or more variables. Changing the subject of a formula means isolating a specific variable on one side of the equation. This is done by applying inverse operations in a specific order to eliminate all other variables and isolate the desired one.

We have the formula as;

[tex]A = \pi R^2 - \pi r^2\\A = \pi(R^2 - r^2)\\A/ \pi = R^2 - r^2\\R^2 = A/ \pi + r^2\\R = \sqrt{} A/ \pi + √r^2\\R = \sqrt{} A/ \pi + r[/tex]

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The required direction of the inequality is below the line.

If you're not graphing a vertical line, the inequality's sign will tell you which half-plane to shade. Shade above the line when using the  or > sign. Shade below the line if the symbol or is used. Larger solutions are to the right and smaller solutions are to the left, respectively, for a vertical line.

To check the direction, we will take points from the graph and put in the given inequality.

At point (2, 0)

y ≤ x/4 - 3

0 ≤ 2/4 - 3

0 ≤ -25

It is false

Again

At (2, -4)

y ≤ x/4 - 3

-4 ≤ 2/4 - 3

-4 ≤ -2.5

It is true,

Thus, we can say the direction of the inequality is below the line.

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a = amount invested at 4%

how much is 4% of "a"? (4/100) * "a", namely 0.04a.

b = amount invested at 5%

how much is 5% of "b"? (5/100) * "b", namely 0.05b.

we know the total amount invested is 5550, so whatever "a" and "b" might be, we know that a + b = 5550.

we also know that the yielded amount in interest is 259, so if we simply add their interest, that'd be 0.04a + 0.05b = 259.

[tex]a+b=5550\implies b=5550-a \\\\[-0.35em] ~\dotfill\\\\ 0.04a+0.05b=259\implies \stackrel{\textit{substituting from above}}{0.04a+0.05(5550-a)~~ = ~~259} \\\\\\ 0.04a+277.5-0.05a=259\implies 277.5-0.01a=259\implies -0.01a=-18.5 \\\\\\ a=\cfrac{-18.5}{-0.01}\implies \boxed{a=1850}\hspace{9em}\stackrel{ 5550~~ - ~~1850 }{\boxed{b=3700}}[/tex]

Answer: 18.75

Step-by-step explanation:

small greeting cards = 1.80

large greeting cards = 3.85

1.80 x 4 = 7.20

3.85 x 3 = 11.55

7.20 + 11.55 = 18.75

We can see that the percentage is 20%, then 20 percent of the students got a 10.

To find this, we need to use the formula:

Percentage = 100%(number of students with a 10)/(total number of students)

Here we know the that the revelant numbers are:

Number of students with a 10 = 12

total number of students in the class = 60

Replacing that in the formula we will get the percentage:

P  = 100%*(12/60)

P = 20%

So 20% of the total number of students got a 10.

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to get the equation of any straight line, we simply need two points off of it, let's use those two in the picture below.

[tex](\stackrel{x_1}{-4}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{-1}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{-1}-\stackrel{y1}{4}}}{\underset{\textit{\large run}} {\underset{x_2}{0}-\underset{x_1}{(-4)}}} \implies \cfrac{-5}{0 +4} \implies \cfrac{ -5 }{ 4 } \implies - \cfrac{ 5 }{ 4 }[/tex]

[tex]\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{4}=\stackrel{m}{- \cfrac{ 5 }{ 4 }}(x-\stackrel{x_1}{(-4)}) \implies y -4 = - \cfrac{ 5 }{ 4 } ( x +4) \\\\\\ y-4=- \cfrac{ 5 }{ 4 }x-5\implies {\Large \begin{array}{llll} y=- \cfrac{ 5 }{ 4 }x-1 \end{array}}[/tex]