A chef has less than 25.8 pounds of rice. She takes out 5 pounds to use during the day and stores the rest in containers. Each container has 2.5 pounds of rice. Let x represent the number of containers. What is the maximum number of containers the chef will be able to fill with rice? 13 12 8 7

Answer:

[tex]\left \{ {{n+q=28} \atop {n(0.05)+q(0.25)=4}} \right.[/tex]

And the pair [tex](n,q)=(15,13)[/tex] is the solution.

Step-by-step explanation:

We know that Yolanda paid for her movie ticket using 28 coins. She only paid with nickels and quarters.

We also know that the ticket cost $4.

We need to form a linear equation system that solves this problem.

We have the following variables :

n : number of nickels

q : number of quarters

We know that she used 28 coins ⇒ the number of nickels plus the number of quarters must be equal to 28.

We have our first equation :

[tex]n+q=28[/tex] (I)

For the second equation we need to use the ticket price information.

We know that the ticket cost $4 and she only paid with nickels and quarters.

Therefore we can write the following equation that relates the variable ''n'' and the variable ''q'' :

[tex]n(0.05)+q(0.25)=4[/tex] (II)

This equation represents that the number of nickels ''n'' per its value plus the number of quarters ''q'' per its value is equal to $4 that it is the value of the movie ticket.

With (I) and (II) we form the linear equation system :

[tex]\left \{ {{n+q=28} \atop {n(0.05)+q(0.25)=4}} \right.[/tex]

This linear equation system can be used to find the value of ''n'' and ''q''.

For example, in equation (I)

[tex]n+q=28[/tex]

we can solve it in terms of ''n'' :

[tex]n=28-q[/tex] (III)

If we use (III) in (II) :

[tex](28-q)(0.05)+q(0.25)=4[/tex]

[tex]1.4-(0.05)q+(0.25)q=4[/tex]

[tex]q(0.2)=2.6[/tex]

[tex]q=\frac{2.6}{0.2}=13[/tex]

[tex]q=13[/tex]

Now replacing this value of q in (III) :

[tex]n=28-13=15[/tex]

[tex]n=15[/tex]

We find that Yoland used 15 nickels and 13 quarters to paid the movie ticket .

Answer:

B. 2 points

Step-by-step explanation:

1 point is a dot in space, 2 can make a line

Answer:

{-5, 3}

Step-by-step explanation:

7 is a factor common to all three terms of this polynomial function.  Thus, after the first attempt to factor it, we have 7(x^2 + 2x - 15).  The quadratic factors into (x + 5)(x - 3).  Setting these = to zero individually, we get

x + 5 = 0, or x = -5, and

x - 3 + 0, or x = 3

Thus, the zeros of the given f(x) =7x2 + 14x − 105 are {-5, 3}.

Answer:

c!

Step-by-step explanation:

Answer:

Option [tex]42.1\ in^{2}[/tex]

Step-by-step explanation:

we know that

The area of the region that is inside the square and outside the circle is equal to the area of the square minus the area of the circle

see the attached figure to better understand the problem

Step 1

Find the area of the square

Remember that

The area of the square is

[tex]A=b^{2}[/tex]

where

b is the length side of the square

we have

[tex]b=14\ in[/tex]

substitute

[tex]A=14^{2}=196\ in^{2}[/tex]

Step 2

Find the area of the circle

Remember that

The area of the circle is equal to

[tex]A=\pi r^{2}[/tex]

we have

[tex]r=14/2=7\ in[/tex]

substitute

[tex]A=\pi(7^{2})=153.9\ in^{2}[/tex]

Step 3

Find the area of the region

[tex]196\ in^{2}-153.9\ in^{2}=42.1\ in^{2}[/tex]

Answer:

k=4

Step-by-step explanation:

Given

A line x=7

The line has only y-intercept which means that it is a vertical line.

As the two points are

A(10,2) and B(k,2)

The y-intercepts of both the points are same i.e. y=2 which means that the line fomed by joining A and B is a horizontal line.  

As x=7 is a vertical line and bisects the horizontal line, the mid-point of line AB will be the point where first line (x=7) bisects the second which is (7,2).

The x-coordinate of mid-point is 7 because the vertical line has 7 as x-intercept and y-intercept is two because the horizontal line has y-intercept 2.

So,

Mid-point=((10+k)/2  ,2)=(7,2)

Comparing the x-coordinate

(10+k)/2=7

10+k=7*2  

10+k=14

k=14-10

k=4

When both the base and height of a shape are multiplied by x, the area increases by a factor of x^2. Therefore, the area of Sail B is x^2 times greater than that of Sail A.

The area of a shape with a base and height is commonly found using the formula Area = 1/2 * Base * Height. If the base and the height of Sail B are x times greater than those of Sail A, then both the base and height of Sail B are multiplied by x, resulting in an area that is x^2 (x squared) times greater. This is because, in the formula, base and height are multiplied together, so if both are multiplied by x, the overall multiplication is by x*x, or x^2. Therefore, the area of Sail B is x^2 times greater than the area of Sail A.

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What is the measure of BC

Solution:

As per Inscribed Angle Theorem: If an angle is inscribed in a circle, then the measure of the intercepted arc is twice the value of the measure of the inscribed angle.

So, as we can see in the diagram the angle inscribed in the circle is 65°.

So, the measure of intercepted arc BC is twice ∠BDC

Or, BC=2*65°

BC=130° Answer

The dimensions of a rectangle with an area of 221 cm² and a perimeter of 60 cm are found by solving a system of equations derived from the definitions of area and perimeter. The dimensions are 13 cm by 17 cm, or equivalently, 17 cm by 13 cm.

To find the dimensions of a rectangle with an area of 221 cm2 and a perimeter of 60 cm, we will let the length be x and the width be y. The area of a rectangle is found by multiplying the length and width, so we have the equation x * y = 221. The perimeter is twice the sum of the length and width, so we have 2x + 2y = 60, which simplifies to x + y = 30.

To solve these equations, divide the perimeter equation by 2 to find y = 30 - x. Substituting this into the area equation gives x(30 - x) = 221. Expanding this and bringing all terms to one side provides a quadratic equation: x2 - 30x + 221 = 0. Solving this quadratic equation by factoring or using the quadratic formula gives the dimensions of the rectangle.

The solutions to the quadratic equation are x = 13 and x = 17. Since x and y are interchangeable as length and width, the two sets of possible dimensions for the rectangle are 13 cm by 17 cm and 17 cm by 13 cm.

The dimensions of the rectangle are [tex]17\ cm , 13\ cm[/tex]

To find the dimensions [tex]\( l \)[/tex] (length) and [tex]\( w \)[/tex] (width) of the rectangle given its area and perimeter, we start with the following equations:

1. Area equation:

[tex]\[l \times w = 221\][/tex]

2. Perimeter equation:

[tex]\[ 2l + 2w = 60 \][/tex]

Step-by-Step Solution:

The dimensions of the rectangle are [tex]{{17, 13} \) cm.[/tex]

From the perimeter equation, divide everything by [tex]2[/tex] to simplify:

[tex]\[l + w = 30\][/tex]

Now we have a system of equations:

[tex]\[\begincases}l \times w = 221 \\l + w = 30\end{cases}\][/tex]

Let's solve this system using substitution or elimination:

From [tex]\( l + w = 30 \)[/tex], we can express [tex]\( w \)[/tex] in terms of [tex]\( l \)[/tex]

[tex]\[w = 30 - l\][/tex]

Substitute [tex]\( w = 30 - l \)[/tex] into the area equation:

[tex]\[l \times (30 - l) = 221\][/tex]

[tex]\[30l - l^2 = 221\][/tex]

Rearrange this equation to form a quadratic equation:

[tex]\[l^2 - 30l + 221 = 0\][/tex]

Now, solve this quadratic equation using the quadratic formula [tex]\( l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -30 \), and \( c = 221 \)[/tex]

[tex]\[l = \frac{-(-30) \pm \sqrt{(-30)^2 - 4 \times 1 \times 221}}{2 \times 1}\][/tex]

[tex]\[l = \frac{30 \pm \sqrt{900 - 884}}{2}\][/tex]

[tex]\[l = \frac{30 \pm \sqrt{16}}{2}\][/tex]

[tex]\[l = \frac{30 \pm 4}{2}\][/tex]

Calculate both possible values of [tex]\( l \)[/tex]

[tex]\[l = \frac{30 + 4}{2} = 17 \quad \text{or} \quad l = \frac{30 - 4}{2} = 13\][/tex]

Corresponding values of \( w \)

[tex]If\ \( l = 17 \), then \( w = 30 - 17 = 13 \)[/tex]

[tex]If\ \( l = 13 \), then \( w = 30 - 13 = 17 \)[/tex]

Therefore, the dimensions of the rectangle are [tex]\( 17 \) cm[/tex] by [tex]\( 13 \) cm.[/tex]

Verification:

[tex]Area: \( 17 \times 13 = 221 \) cm\(^2\)[/tex]

[tex]Perimeter: \( 2 \times (17 + 13) = 2 \times 30 = 60 \) cm[/tex]

Both conditions match the given area and perimeter, confirming that the dimensions are correct.

divide number of apples by how many fit I a basket:

285 / 37 = 7.70

 so he will need 8 baskets

Final answer:

To find the total distance to town, the distance walked (1.5 miles) and run (1.75 miles) were calculated, and it was found that the student is 3.25 miles away from home. This is halfway to town, so the total distance to town is 6.5 miles.

Explanation:

To solve the problem, we will calculate the distance traveled by walking and running and then use these to find the total distance to town.

Step 1: Calculate the distance walked

First, we convert the walking time into hours since the speed is given in miles per hour. Half an hour of walking at 3 miles per hour will cover:

Distance walked = Speed × Time = 3 mph × 0.5 hours = 1.5 miles

Step 2: Calculate the distance run

Since 15 minutes is a quarter of an hour, we can calculate the distance run at 7 miles per hour as:

Distance run = Speed × Time = 7 mph × 0.25 hours = 1.75 miles

Step 3: Determine the total half distance

Now, we combine the distances to find the halfway point to town:

Total half distance = Distance walked + Distance run = 1.5 miles + 1.75 miles = 3.25 miles

Step 4: Find the full distance to town

Since this is half the distance, the full distance to town is:

Full distance = Total half distance × 2 = 3.25 miles × 2 = 6.5 miles

Conclusion

Therefore, the total distance from the student's house to town is 6.5 miles.

Final answer:

The total distance from the house to town is found by doubling the sum of the distances walked and run, which were 1.5 miles and 1.75 miles, respectively. The full distance to town is 6.5 miles.

Explanation:

To solve for the distance from your house to town, we will use the information provided about your walking and running times and speeds. Since you walked for half an hour (0.5 hours) at a speed of 3 miles per hour and ran for 15 minutes (0.25 hours) at a speed of 7 miles per hour, and this brought you halfway to town, we can set up the following equation:

Distance walked = Speed walked × Time walked
= 3 mph × 0.5 hours
= 1.5 miles

Distance run = Speed run × Time run
= 7 mph × 0.25 hours
= 1.75 miles

The total distance you traveled to get halfway to town is the sum of the distance walked and the distance run:

Total distance halfway = Distance walked + Distance run
= 1.5 miles + 1.75 miles
= 3.25 miles

Since 3.25 miles is only halfway to town, the full distance to town is twice that amount:

Full distance to town = 2 × Total distance halfway
= 2 × 3.25 miles
= 6.5 miles

Therefore, the total distance from your house to town is 6.5 miles.

Answer:the correct answer is 71142 just multiply 334 and 213

Step-by-step explanation:

L=2+4W
W=L/2
f=3d

hope this is the right question????

GDP calculations should include 'final goods and services' and 'new products' to reflect a nation's current economic activity accurately while preventing double counting.

The question pertains to what should be included when calculating the Gross Domestic Product (GDP). The options given were a. final goods and services, b. intermediate goods and services, c. last year's products, and d. new products. According to the principles of economics, GDP is the market value of all final goods and services produced within a country in a given period of time.

It is crucial to understand that GDP calculations include only new products that are produced within the current year to reflect the nation's current economic activity. This method helps statisticians avoid the mistake of double counting, which would inflate GDP figures inaccurately. Consequently, the correct selections are a. final goods and services and d. new products.