Joe bought g gallons of gasoline for $2.85 per gallon and c cans of oil for $3.15 per can. What expression can be used to determine the total amount Joe spent on gasoline and oil?

Answer:

2/5 < 0.42 < 43% < 3/7

Step-by-step explanation:

Let's convert them all to decimals:

43% = 0.43

2/5 = 0.4

3/7 = 0.428571...

0.42 = 0.42

Now we can arrange them in ascending order:

0.4

0.42

0.43

0.428571...

The correct statement is: "The line whose equation is 3x-5y=4 is dilated by a scale factor of [tex]y= (\frac{5}{3} )x[/tex] centered at the origin, and the equation of the dilated line is y= (\frac{5}{3} )x

When a line is dilated by a scale factor of k centered at the origin, the equation of the dilated line is given by y = kx, if the original line passes through the origin. If the original line does not pass through the origin, then the equation of the dilated line is obtained by finding the intersection point of the original line with the line passing through the origin and the point of intersection of the original line with the x-axis, dilating this intersection point by the scale factor k, and then finding the equation of the line passing through this dilated point and the origin.

In this case, the equation of the original line is 3x - 5y = 4. To find the intersection point of this line with the x-axis, we set y = 0 and solve for x:

3x - 5(0) = 4
3x = 4
[tex]x = \frac{4}{3}[/tex]

Therefore, the intersection point of the original line with the x-axis is (4/3, 0). Dilating this point by a scale factor of 5/3 centered at the origin, we obtain the dilated point:

[tex](\frac{5}{3} ) (\frac{4}{3},0) = (\frac{20}{9},0)[/tex]

The equation of the dilated line passing through this point and the origin is given by [tex]y= (\frac{5}{3} )x[/tex]. Therefore, the correct statement is: "The line whose equation is 3x-5y=4 is dilated by a scale factor of [tex]\frac{5}{3}[/tex] centered at the origin, and the equation of the dilated line is [tex]y= (\frac{5}{3} )x[/tex]."

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The rate at which the current is changing is -1/32 amperes per second (A/s).

To find the rate at which the current is changing, we will use the given information and apply the differentiation rules. The terms we will use in the answer are voltage (V), resistance (R), current (I), and rate of change.

Given the formula for current: I = V/R
We have V = 10 volts (constant) and dR/dt = 0.2 ohms/second.

We need to find dI/dt, the rate at which the current is changing. To do this, we differentiate the formula for current with respect to time (t):

[tex]dI/dt = d(V/R)/dt[/tex]
Since V is constant, its derivative with respect to time is 0.

dI/dt = -(V * dR/dt) / R^2 (using the chain rule for differentiation)

Now, substitute the given values:

[tex]dI/dt = -(10 * 0.2) / 8^2[/tex]
[tex]dI/dt = -2 / 64[/tex]
[tex]dI/dt = -1/32 A/s[/tex]

The rate at which the current is changing is -1/32 amperes per second (A/s).

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For the expressions  3x+4 –2x, and 5x+2x+x the simplified expression after combining like terms is x+4 and 8x.

The given expressions are  3x+4 –2x, and 5x+2x+x

We have to simplify these expressions by combining the like terms

For the expression 3x+4 –2x

We have to combine like terms

x+4

Now for expression  5x+2x+x

Combine the like terms to get

8x

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The correct choice is: C. The absolute maximum is at x = 1050, and there is no absolute minimum.

To find the absolute extrema of the function f(x) = -0.002x^2 + 4.2x - 50 over the interval (-∞, ∞), we need to find the critical points and then determine if there's a maximum or minimum at each point.

Step 1: Find the derivative of the function f(x) with respect to x. f'(x) = -0.004x + 4.2

Step 2: Set the derivative equal to zero and solve for x. -0.004x + 4.2 = 0 x = 1050

Step 3: Since we have only one critical point, we need to determine if it's a maximum or a minimum. To do this, we can use the second derivative test.

Step 4: Find the second derivative of the function f(x) with respect to x. f''(x) = -0.004

Step 5: Since the second derivative is negative (f''(x) = -0.004 < 0), the critical point x = 1050 corresponds to an absolute maximum. Step 6: Calculate the value of the function f(x) at x = 1050. f(1050) = -0.002(1050)^2 + 4.2(1050) - 50 = 2150

Thus, the absolute maximum is at x = 1050, and the value is 2150. Since the function is a parabola with the "mouth" facing downwards, there is no absolute minimum.

The correct choice is: C. The absolute maximum is at x = 1050, and there is no absolute minimum.

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A graph of the points that represent this data are shown on the coordinate plane attached below.

In this scenario, the week number would be plotted on the x-axis (x-coordinate) of the scatter plot while the amount of money (in dollars) would be plotted on the y-axis (y-coordinate) of the scatter plot through the use of Microsoft Excel.

On the Excel worksheet, you should right click on any data point on the scatter plot, select format trend line, and then tick the box to display an equation for the line of best fit (trend line) on the scatter plot.

From the scatter plot (see attachment) which models the relationship between the week number and the amount of money (in dollars), a linear equation for the line of best fit is as follows:

y = 1.19x + 1.05

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We have found a value of b that makes y(t) = [tex]e^2t[/tex] [1; 4; -1/2] a solution of y′ = [4 1 3; 2 3 3; −2 −1 −1]y. To check if y(t) is a solution of y′ = Ay, we need to substitute it into the differential equation and see if it holds.

Let's start by finding y′:

y′(t) = [[tex]2e^2t, 8e^2t, 4be^2t[/tex]]

Now, let's find Ay:

Ay = [4 1 3; 2 3 3; −2 −1 −1] [1; 4; b] = [4+4b; 14; -5-b]

We want y(t) = e^2t [1; 4; b] to satisfy y′ = Ay, so we set them equal:

y′ = Ay

[[tex]2e^2t; 8e^2t; 4be^2t] = [4+4b; 14; -5-b] e^2t[/tex] [1; 4; b]

Expanding this equation, we get:

[tex]2e^2t[/tex]= (4+4b)[tex]e^2t[/tex]

[tex]8e^2t[/tex] = 14 [tex]e^2t[/tex]

[tex]4be^2t[/tex]= (-5-b) [tex]e^2t[/tex]

The second equation is always true, so we can ignore it. For the first equation, we can cancel out [tex]e^2t[/tex] on both sides to get:

2 = 4+4b

Solving for b, we get:

b = -1/2

Finally, we can substitute b = -1/2 back into the third equation to check if it holds:

4be^2t = (-5-b) [tex]e^2t[/tex]

-2e^2t = (-5 + 1/2)[tex]e^2t[/tex]

This equation is true, so we have found a value of b that makes y(t) = [tex]e^2t[/tex] [1; 4; -1/2] a solution of y′ = [4 1 3; 2 3 3; −2 −1 −1]y.

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Answer:

Step-by-step explanation:

625

Answer:

2

Step-by-step explanation:

The K and N are the trapeziums and the two lines opposite to them go in a parallel line

Josue and Pho brought 3 1/5 liters of lemonade altogether

Josue and Pho brought 1 3/5 liters of lemonade each, so the total amount of lemonade they brought is:

1 3/5 + 1 3/5 = 3 1/5

To add the two mixed numbers, we first need to find a common denominator. In this case, the common denominator is 5. Then we convert both mixed numbers into fractions with a denominator of 5:

1 3/5 = (5 × 1 + 3) / 5 = 8/5

1 3/5 = (5 × 1 + 3) / 5 = 8/5

Now we can add the fractions:

8/5 + 8/5 = (8 + 8) / 5 = 16/5

Finally, we can convert the fraction back to a mixed number:

16/5 = 3 1/5

Therefore, Josue and Pho brought 3 1/5 liters of lemonade altogether.

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The equation of the hyperbola is -(x + 5)²/71 + (y - 2)² = -71

We can also find the distance between the center and the given focus, which is the distance between (-5, 2) and (-5 - √29, 2):

d = |-5 - (-5 - √29)| = √29

Substituting in the known values, we get:

c² = a² + b²

(√29)² = (10)² + b²

29 = 100 + b²

b² = -71

(x - h)²/a² - (y - k)²/b² = 1

where (h, k) is the center of the hyperbola.

Substituting in the known values, we get:

(x + 5)²/100 - (y - 2)²/-71 = 1

Multiplying both sides by -71, we get:

-(x + 5)²/71 + (y - 2)²/1 = -71/1

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(a) y(x) = c1 e^(-4x/3) cos(2x) + c2 e^(-4x/3) sin(2x), stable node at the origin;

(b) y(x) = c1 e^(2x) + c2 e^(-4x), unstable node at the origin;

(c) y(x) = c1 e^(-x/22) cos(sqrt(119)x/22) + c2 e^(-x/22) sin(sqrt(119)x/22), stable node at the origin;

(d) y(x) = c1 e^(sqrt(3)x/2) + c2 e^(-sqrt(3)x/2), unstable saddle at the origin;

(e) y(x) = c1 e^x cos(2x) + c2 e^x sin(2x), stable spiral at the origin;

(f) y(x) = c1 cos(3x/2) + c2 sin(3x/2), stable limit cycle around the origin;

(g) y(x) = c1 e^(-x/3) + c2 e^(-x), stable node at the origin.

(a) The characteristic equation is 2r^2 + 9r + 4 = 0, with roots r1 = -4/3 and r2 = -1/2. The general solution is y(x) = c1 e^(-4x/3) cos(2x) + c2 e^(-4x/3) sin(2x). The equilibrium at the origin is a stable node since both eigenvalues have negative real parts.

(b) The characteristic equation is r^2 + 2r - 8 = 0, with roots r1 = 2 and r2 = -4. The general solution is y(x) = c1 e^(2x) + c2 e^(-4x). The equilibrium at the origin is an unstable node since both eigenvalues have positive real parts.

(c) The characteristic equation is 44r^2 - 12r + 5 = 0, with roots r1 = (3 + sqrt(119))/22 and r2 = (3 - sqrt(119))/22. The general solution is y(x) = c1 e^(-x/22) cos(sqrt(119)x/22) + c2 e^(-x/22) sin(sqrt(119)x/22). The equilibrium at the origin is a stable node since both eigenvalues have negative real parts.

(d) The characteristic equation is 2r^2 - 3 = 0, with roots r1 = sqrt(3)/2 and r2 = -sqrt(3)/2. The general solution is y(x) = c1 e^(sqrt(3)x/2) + c2 e^(-sqrt(3)x/2). The equilibrium at the origin is an unstable saddle since the eigenvalues have opposite signs.

(e) The characteristic equation is r^2 - 2r + 5 = 0, with roots r1 = 1 + 2i and r2 = 1 - 2i. The general solution is y(x) = c1 e^x cos(2x) + c2 e^x sin(2x). The equilibrium at the origin is a stable spiral since both eigenvalues have negative real parts and non-zero imaginary parts.

(f) The characteristic equation is 4r^2 + 9 = 0, with roots r1 = 3i/2 and r2 = -3i/2. The general solution y(x) = c1 cos(3x/2) + c2 sin(3x/2), stable limit cycle around the origin.

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You need the unit price $0.40/mile to find the cost of a 47-mile taxi ride.

Use the unit price to find the cost of a 47-mile taxi ride

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A t-test with a level of significance of 0.05 results in a test statistic of -2.09, indicating a significant difference between the population mean for the math scores and the population mean for the EBRW scores.

To test for a difference between the population mean for the math scores and the population mean for the ebrw scores, we can conduct a two-sample t-test.

Using a calculator or software, we can find that the sample mean for math scores is 520.5 and the sample mean for ebrw scores is 485.5.

The sample size is n = 12 for both groups.

The sample standard deviation for math scores is s1 = 48.50 and for ebrw scores is s2 = 87.63.

Using a level of significance of 0.05, and assuming unequal variances, we can find the test statistic as:

t = (520.5 - 485.5) / sqrt(([tex]48.50^2/12[/tex]) + ([tex]87.63^2/12[/tex]))

t = 0.851

Rounding to two decimal places, the test statistic is 0.85.

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To find the derivative of the function h(p) = -5/(p^2+1), we will use the quotient rule:

h'(p) = [(-5)'(p^2+1) - (-5)(p^2+1)'] / (p^2+1)^2

Simplifying this expression, we get:

h'(p) = (10p) / (p^2+1)^2

To find the values of p such that h'(p) = 0, we will set the numerator equal to 0 and solve for p:

10p = 0

p = 0

Therefore, h'(p) = 0 when p = 0.

To find the values of p in the domain of h such that h'(p) does not exist, we need to find the values of p where the denominator of h'(p) becomes 0:

p^2+1 = 0

This equation has no real solutions, so there are no values of p in the domain of h such that h'(p) does not exist. Therefore, we enter DE (does not exist).

To find the critical numbers of the function, we need to find the values of p where h'(p) = 0 or h'(p) does not exist. We have already found that h'(p) = 0 when p = 0, and we have determined that h'(p) does not exist for any values of p in the domain of h. Therefore, the only critical number of the function is p = 0.
Let's first find the derivative of the given function, h(p) = (p - 5)/(p^2 + 1).

Using the quotient rule, h'(p) = [(p^2 + 1)(1) - (p - 5)(2p)]/((p^2 + 1)^2).

Simplifying, h'(p) = (p^2 + 1 - 2p^2 + 10p)/((p^2 + 1)^2) = (-p^2 + 10p + 1)/((p^2 + 1)^2).

To find the values of p such that h'(p) = 0, set the numerator of h'(p) equal to zero:

-p^2 + 10p + 1 = 0.

This is a quadratic equation, but it does not have any real solutions. Therefore, there are no values of p for which h'(p) = 0, so the answer is DNE.

To find the values of p where h'(p) does not exist, we look for where the denominator is zero:

(p^2 + 1)^2 = 0.

However, this equation has no real solutions, as (p^2 + 1) is always positive. Therefore, there are no values of p for which h'(p) does not exist, so the answer is DE.

Since there are no values of p for which h'(p) = 0 and no values of p for which h'(p) does not exist, there are no critical numbers of the function. The answer is DNE.

Your answer:
h(p) = (p - 5)/(p^2 + 1)
h'(p) = (-p^2 + 10p + 1)/((p^2 + 1)^2)
p (h'(p) = 0) = DNE
p (h'(p) does not exist) = DE
Critical numbers = DNE

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Number of cars that coasted more than 10.75 feet = 1

To solve this problem, you need to compare the distance each car coasted to 10.75 feet, which is the threshold for determining whether a car coasted more or less than 10.75 feet.

Manuel's car coasted 10 feet, which is less than 10.75 feet, so it did not coast more than 10.75 feet.

Richard's car coasted 10.5 feet, which is also less than 10.75 feet, so it did not coast more than 10.75 feet either.

Diego's car coasted 102 feet, which is more than 10.75 feet. Therefore, only one car coasted more than 10.75 feet, and the answer is 1.

So the answer is:

Number of cars that coasted more than 10.75 feet = 1

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Answer:

2.25

Step-by-step explanation:

45 nickels

45*5=225

225 cents

2.25

The value of 45 nickels in decimal number can be 2.25.

In the decimal system, each digit's value depends on its position or place value within the number.

A nickel is worth 0.05 dollars.

To find the value of 45 nickels, multiply the number of nickels by the value of each nickel:

So, Value = Number of nickels × Value of each nickel

                = 45 × 0.05

                = 2.25

Therefore, the value of 45 nickels is $2.25 as a decimal number.

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To create a function that combines the two scenarios, we need to add the amount of money you earn from babysitting to the amount of money you have in your suitcase. We can represent this with the following function:

f(x) = s(x) + a(x)

Where f(x) represents the total amount of money you have after x hours of babysitting. We substitute s(x) with the given function, s(x) = 325, and a(x) with the given function, a(x) = 5(x-2):

f(x) = 325 + 5(x-2)

Simplifying this expression, we can distribute the 5 to get:

f(x) = 325 + 5x - 10

And then combine the constant terms:

f(x) = 315 + 5x

So the function that combines the two scenarios is f(x) = 315 + 5x. This function gives you the total amount of money you will have after x hours of babysitting and taking into account the initial amount of money you have in your suitcase.

In summary, to create a function that combines the two scenarios, we simply add the amount of money earned from babysitting to the initial amount of money in the suitcase. The function f(x) = 315 + 5x represents this total amount of money.

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The theoretical probability of the spinner not landing on yellow, would be 75 %.

In order to calculate the likelihood of the spinner not landing on yellow, it is necessary to initially identify the quantity of non-yellow partitions and subsequently divide this by the full tally of sections. The spinner comprises a total of 8 individual segments.

Of these, two (i.e., sections 2 and 3) are colored in shades of yellow, hence totaling two yellow sectors. This leaves a further six compartments - numbered 1, 4, 5, 6, 7 and 8, that do not fall into the category of "yellow."

The probability is therefore :

= ( Number of not yellow sections ) / ( Total number of sections )

= 6 / 8

= 3 / 4

= 75 %

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Step-by-step explanation:

[tex] \frac{1}{( {x}^{2} + 1) } = \frac{u}{v} [/tex]

u = 1

u' = 0

v = x² + 1

v' = 2x

[tex] \frac{1}{ ({x}^{2} + 1)} \\ = \frac{u'v - v'u}{ {v}^{2} } \\ = \frac{0 - (2x \times 1)}{ {( {x}^{2} + 1)}^{2} } \\ = - \frac{2x}{ { ({x}^{2} + 1) }^{2} } [/tex]

Answer:

Sure, here's the solution to the word problem:

Jack has $10 in his lunch account and plans to spend $2 a week on snacks. To find out how long it will take his lunch account to reach zero, we can divide the total amount of money in his account by the amount he spends each week.

```

$10 / $2 = 5 weeks

```

Therefore, it will take Jack 5 weeks to spend all of the money in his lunch account.

Here's another way to solve the problem:

We can also set up an equation to represent the situation. Let x be the number of weeks it takes Jack's lunch account to reach zero. We know that Jack starts with $10 and spends $2 each week, so we can write the equation:

```

$10 - $2x = 0

```

Solving for x, we get:

```

x = 5

```

Therefore, it will take Jack 5 weeks to spend all of the money in his lunch account.

Answer:

in 5 weeks he will have 0$ in his account

Step-by-step explanation:

In circle P, if m(QR) = 80 , and m(QRT) = 39 , m(QPR) = 39 and m(PT) = 78

Based on the information given, we know that:

- m(QR) = 80 (this is the measure of arc QR)
- m(QRT) = 39 (this is the measure of angle QRT)

To find the other measures, we can use the following formulas:

- The measure of a central angle is equal to the measure of its intercepted arc
- The measure of an inscribed angle is half the measure of its intercepted arc

Using these formulas, we can find the measure of angle QPR and the measure of arc PT as follows:

- m(QPR) = m(QRT) = 39 (since angle QRT and angle QPR intercept the same arc QR)
- m(PT) = 2 * m(QRT) = 78 (since angle QRT and angle PQT intercept the same arc PT, and the measure of an inscribed angle is half the measure of its intercepted arc)

So the final answers are:

- m(QR) = 80
- m(QRT) = 39
- m(QPR) = 39
- m(PT) = 78

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Answer:

  a) multiply by cos²/cos², move sin/cos inside parentheses, simplify

  d) multiply by (cot+cos); use cot=cos·csc, csc²-1=cot² in the denominator

Step-by-step explanation:

You want to prove the identities ...

Usually, we want to prove a trig identity by providing the steps that transforms one side of the identity to the expression on the other side. Here, each of these identity expressions can be simplified, so it is actually much easier to simplify both expressions to one that is common.

We are going to use s=sin(x), c=cos(x), (s/c) = tan(x), and (c/s) = cot(x) to reduce the amount of writing we have to do.

  [tex]s^2\left(\dfrac{c}{s}+1\right)^2=c^2\left(\dfrac{s}{c}+1\right)^2\qquad\text{given}\\\\\\\dfrac{s^2(c+s)^2}{s^2}=\dfrac{c^2(s+c)^2}{c^2}\qquad\text{use common denominator}\\\\\\(c+s)^2=(c+s)^2\qquad\text{cancel common factors; Q.E.D.}[/tex]

Using the same substitutions as above, we have ...

  [tex]\dfrac{c(c/s)}{(c/s)-c}=\dfrac{(c/s)+c}{c(c/s)}\qquad\text{given}\\\\\\\dfrac{c^2}{c(1-s)}=\dfrac{c(1+s)}{c^2}\qquad\text{multiply num, den by s}\\\\\\\dfrac{c(1+s)}{(1-s)(1+s)}=\dfrac{c(1+s)}{c^2}\\\\\\\dfrac{c(1+s)}{1-s^2}=\dfrac{c(1+s)}{c^2}\\\\\\\dfrac{c(1+s)}{c^2}=\dfrac{c(1+s)}{c^2}\qquad\text{Q.E.D.}[/tex]

__

Additional comment

The key transformation in (d) is multiplying numerator and denominator by (1+sin(x)). You can probably prove the identity just by doing that on the left side, then rearranging the result to make it look like the right side.

For (a), the key transformation seems to be multiplying by cos²(x)/cos²(x) and rearranging.

Sometimes it seems to take several tries before the simplest method of getting from here to there becomes apparent. The transformations described in the top "Answer" section may be simpler than those shown in the "Step-by-step" section.

the MAXIMUM area that can be enclosed is 900 m²

To find the maximum area that can be enclosed, we need to find the vertex of the parabolic function A(x) = -4x^2 + 120x. The vertex represents the maximum point on the parabola.

The x-coordinate of the vertex can be found using the formula x = -b/2a, where a is the coefficient of the x^2 term and b is the coefficient of the x term. In this case, a = -4 and b = 120, so x = -120/(2*(-4)) = 15.

To find the y-coordinate of the vertex, we can substitute x = 15 into the function: A(15) = -4(15)^2 + 120(15) = 900. Therefore, the maximum area that can be enclosed is 900 square meters.

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Answer:

Step-by-step explanation:

The total balance of the two investment accounts at the end of ten years will be $16,564.08. To calculate the total balance of the two accounts at the end of ten years,

we need to use the formulas for simple interest and compound interest.

For Account I, the simple interest formula is:

I = Prt

where I is the interest earned, P is the principal (the amount deposited), r is the annual interest rate as a decimal, and t is the time in years.

Plugging in the values for Account I, we get:

I = (3500)(0.065)(10) = $2,275

So, after ten years, the balance in Account I will be:

B1 = P + I = 3500 + 2275 = $5,775

For Account II, the compound interest formula is:

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

where A is the balance at the end of the time period, P is the principal, r is the annual interest rate as a decimal, n is the number of times the interest is compounded per year, and t is the time in years.

Plugging in the values for Account II, we get:

A = 5900(1 + 0.06/1)^(1*10) = $10,789.08

So, after ten years, the balance in Account II will be $10,789.08.

Therefore, the total balance of the two accounts at the end of ten years will be:

Total balance = Balance in Account I + Balance in Account II

= $5,775 + $10,789.08

= $16,564.08

In summary, by using the formulas for simple interest and compound interest, we can calculate that the total balance of the two investment accounts at the end of ten years will be $16,564.08.

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Answer:

B. The graph has a vertical asymptote at

x = -2.

The statement about the graph of the given rational function that is true is: B. The graph has a vertical asymptote at x = -2.

To understand the graph of the rational function f(x) = (3x - 7) / (x + 2), we need to consider its behavior at various points. First, let's investigate the possibility of asymptotes. Asymptotes are lines that the graph approaches but never touches. There are two types of asymptotes: vertical and horizontal.

A vertical asymptote occurs when the denominator of the rational function becomes zero. In this case, the denominator is (x + 2), so we need to find the value of x that makes it zero. Setting x + 2 = 0 and solving for x, we get x = -2. Therefore, the rational function has a vertical asymptote at x = -2 (option B).

To determine if there is a horizontal asymptote, we need to compare the degrees of the numerator and the denominator. The degree of a term is the highest power of x in that term. In the given rational function, the degree of the numerator is 1 (3x) and the degree of the denominator is also 1 (x). When the degrees are the same, we look at the ratio of the leading coefficients, which are 3 (numerator) and 1 (denominator). The ratio of the leading coefficients is 3/1 = 3.

If the ratio of the leading coefficients is a finite value (not zero or infinity), then the rational function will have a horizontal asymptote. In this case, the horizontal asymptote is y = 3 (option C).

Hence the correct option is (b).

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Answer:

108y^10x^17

Step-by-step explanation:

If Sarah uses 3/4 yard of ribbon to make a hair bow, she will need 6 and 3/4 yards of ribbon to make 9 hair bows.

To find out how many yards of ribbon Sarah will use to make 9 hair bows, we need to multiply the amount of ribbon used for one hair bow (3/4 yard) by the number of hair bows she wants to make (9).

So, the equation we need to use is:

3/4 yard of ribbon per hair bow x 9 hair bows = ? yards of ribbon

To solve for the answer, we can simplify the equation:

3/4 x 9 = 27/4

So Sarah will need 27/4 yards of ribbon to make 9 hair bows.

To convert this fraction to a mixed number, we can divide the numerator (27) by the denominator (4) and write the remainder as a fraction:

27 ÷ 4 = 6 with a remainder of 3

In summary, Sarah will need 6 and 3/4 yards of ribbon to make 9 hair bows, if she uses 3/4 yard of ribbon to make one hair bow.

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Answer: 3

Step-by-step explanation:

Answer:

2hrs 24 mins

Step-by-step explanation:

Ok so let's make this problem a bit simpler by splitting it up.

Tim paints a room in 6 hours.

So, we can also say that she paints 1/6 of that room in 1 hour

Bella paints it in 4 hours

So, we can also say that she paints 1/4 of that room in 1 hour

Now, lets see what we have:

Bella: 1/4 every hour

Tim: 1/6 every hour

The problem states that they are working together, so we need to add the values we have:

1/4 + 1/6

We cannot just add them, we must make them have the same common denominator.

LCD is 12, you can find that by just doing the times tables for 4 and 6 and seeing what number they match on first.

3/12 + 2/12 = 5/12

So, tim and bella working together paint 5/12 of a room in 1 hour.

They paint 5/12 of a room in 60 minutes

They paint 1/12 of the room in 12 minutes(divide both values by 5)

So if they paint 1/12 of the room in 12 minutes, we can multiply both values by 12 to get our answer.

They paint the full room in 144 minutes(12*12).

144 minutes is 2 hours and 24 minutes

Answer: It's correct

Step-by-step explanation: