The manager of a plant nursery bought 120 lb of soil to use for potting plants. the manager can put 5 lb into each of the small pots and 12 lb into each of the large pots. the manager made a sketch to show the line representing the different combinations of pots that can be used. the x-axis represented the number of small pots and the y-axis represented the number of large pots. the manager labeled only the intercepts.which points did the manager label?a.(0, 5)(0, 5) and (12, 0)(12, 0)
b.(0, 10)(0, 10) and (24, 0)(24, 0)
c.(0, 12)(0, 12) and (5, 0)(5, 0)
d.(0, 24)(0, 24) and (10, 0)

Given :

The temperature of a pot of water is 62 F .

The temperature increases  by 20 F per minute when being heated .

To Find :

Write an equation to represent  the linear relationship.

Solution :

Here , the temperature is increasing directly with the time .

Therefore , it is a linear equation .

Let temperature varies with as :

T = at+b   .... equation 1 .

( Here , a and b are constants )

Now , at t= 0 min .

Now , after 1 min temperature will be , T = 62 +20 =82 F

Putting value of a ans b in equation 1 , we get :

Hence , this is the required relation .

The length of the side of the original cube in the given scenario is 5 inches.

A cube is a three-dimensional figure with equal sides.

Given that

The length of the edge of a cube = x

Also, one of the sides is increased by 4 inches, and another side is doubled,

so, now the new dimensions are x, x+4, 2x, and the shape is a rectangular prism.

The area of the rectangular prism = 450 sq. inches

base × height × length = 450 sq. inches

x × (x + 4) × 2x = 450

2x³ + 8x² = 450

2x³ + 8x² - 450 = 0

x³ + 4x² - 225 = 0

Substitute x=5  to see that it is a root of the equation,

5³ + 4(5)² - 225 = 125 + 100 - 225 = 0

So, x - 5 is a factor of the equation,

Now,

x³ + 4x² - 225 = (x−5)(x² + 9x + 45) = 0

The roots of the quadratic equation are imaginary as

D = 9² - 4 × 1 × 45 = 81 - 180 = -99

The discriminant is negative, therefore, the roots are not real.

So, the length of the side of the original cube is 5 inches.

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Answer:

A. Gradually increase in difficulty

Step-by-step explanation:

Option A is the most effective choice because structuring the first few questions of a long calculus exam to gradually increase in difficulty can help students ease into the exam, build confidence, and maintain their concentration. Starting with easier questions allows students to warm up and gain momentum, which can reduce anxiety and increase their focus. This approach aligns with best practices in assessment and educational psychology, as it promotes a smoother transition into more challenging material, ultimately supporting better concentration and performance throughout the exam.

Final answer:

To maintain student concentration during a long exam, the first few questions should gradually increase in difficulty. This approach builds student confidence and eases them into the problem-solving process, potentially reducing test anxiety and encouraging perseverance through harder problems.

Explanation:

To help his students maintain complete concentration throughout their long calculus exam, Mr. Griffin should structure the first few questions to be gradually increase in difficulty. This approach helps students to gain confidence as they successfully solve the initial questions which is likely to carry them through the rest of the exam and maintain their concentration.

Beginning with easier questions allows the students to 'warm up' and transition their mind into the calculus mode. Then, as the questions become increasingly difficult, students are better prepared to tackle them because they've eased into the problem-solving process instead of being hit with the most challenging problems right off the bat. This approach can reduce test anxiety and encourage perseverance through the more difficult problems towards the end of the test.

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The equation gives the value of h in terms of V (volume) and B (base area) for the given equation .

Given equation:

Step 1: Multiply both sides of the equation by 3 to isolate Bh:

Step 2: Divide both sides of the equation by B to solve for h:

So, the equation solved for h is:

This gives you the value of h in terms of V (volume) and B (base area) for the given equation .

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Answer:

h=3v/b

Step-by-step explanation:

Final answer:

The slope of the function y = 2x^2 + 2 at the point x = 3 is 12.

Explanation:

The slope of a function can be determined using the derivative of the function. In this case, the function is y = 2x^2 + 2. Taking the derivative of this function gives us dy/dx = 4x. To find the slope at x = 3, we substitute x = 3 into the derivative: dy/dx = 4(3) = 12.

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