A snow cone holder is a cone with diameter of 3 inches and a height

What is the solution to the inequality shown below?

lesantik [10]

To solve the inequality, you need to isolate/get the variable "p" by itself in the inequality:

5p + 26 < 72 Subtract 26 on both sides

5p + 26 - 26 < 72 - 26

5p < 46 Divide 5 on both sides to get "p" by itself

$\frac{5p}{5}$

$p$

p < 9.2 Your answer is A

8 0

3 years ago

Use long division to determine the quotient of the polynomials. (x3 – 7x – 6) ÷ (x – 4) Which of the following options best desc

Lelechka [254]

Answer:

The correct option is c.

Step-by-step explanation:

The quotient of the polynomials is

$\frac{(x^3-7x-6)}{x-4}$

We need to find the remainder by using long division method.

The dividend of the given expression is

$Dividend=x^3-7x-6$

$Divisor=x-4$

The long division method is shown below.

$\frac{(x^3-7x-6)}{x-4}=(x^2+4x+9)+\frac{30}{x-4}$

From the below attachment it is clear that the quotient has a remainder of 30. Therefore, the divisor is not factor of the dividend.

Therefore the correct option is c.

4 0

2 years ago

Perform a first derivative test on the function f(x)equals2 x cubed plus 3 x squared minus 120 x plus 6; [minus5,8]. Bold

maria [59]

Answer:

a) Critical points

x = 4 and x = -5

b) x = 4 corresponds to a minimum point for the function f(x)

x = - 5 corresponds to a maximum point for the function f(x)

c) The minimum value of f(x) in the interval = -298

The maximum value of f(x) in the interval = 431;

Step-by-step explanation:

f(x) = 2x³ + 3x² - 120x + 6 in the interval [-5, 8]

a) To obtain the critical points, we need to obtain the first derivative of the function with respect to x. Because at critical points of a function, f(x), (df/dx) = 0

f'(x) = (df/dx) = 6x² + 6x - 120 = 0

6x² + 6x - 120 = 0

Solving the quadratic equation,

x = 4 or x = -5

The two critical points, x = 4 and x = -5 are in the interval given [-5, 8] (the interval includes -5 and 8, because it's a closed interval)

b) To investigate the nature of the critical points, we obtain f"(x)

Because f'(x) changes sign at the critical points

If f"(x) > 0, then it's a minimum point and if f"(x) < 0 at c, then it's a maximum point.

f"(x) = (d²f/dx²) = 12x + 6

at critical point x = 4

f"(x) = 12x + 6 = 12(4) + 6 = 54 > 0, hence, x = 4 corresponds to a minimum point.

at critical point x = -5

f"(x) = 12x + 4 = 12(-5) + 4 = -56 < 0, hence, x = -5 corresponds to a maximum point.

c) At x = 4,

f(x) = 2x³ + 3x² - 120x + 6 = 2(4)³ + 3(4)² - 120(4) + 6 = -298

At x = - 5

f(x) = 2x³ + 3x² - 120x + 6 = 2(-5)³ + 3(-5)² - 120(-5) + 6 = 431

7 0

2 years ago

Complete the tables below by using the properties of operations (help)

disa [49]

Answer:

-3(4x+3)+5x (Given) 1/2 (6k+10)+5k (given)

-12x-9+5x (distribution) 3k+5+5k (distribution)

-12x+5x-9 (associative) 3k+5k+5 (associative)

-7x-9 (Combine like terms) 8k+5 (combine like terms)

Step-by-step explanation:

6 0

2 years ago

Function tables with two-step rules Fill in the table using this function rule.<br> help asap

NeTakaya

Answers:

Answer for row one: 1

Answer for row two: 11

Answer for row three: 16

Answer for row four: 36

=========================================

Work Shown:

Whatever the x value is, we multiply by 5 and subtract off 14 to get the corresponding y value. This is following the order of operations PEMDAS

------------

If x = 3, then

y = 5*x - 14

y = 5*3 - 14 ..... note how x is replaced with 3

y = 15 - 14

y = 1

This means that when x = 3, the y value is y = 1.

So 1 goes in the box in the first row.

---------------

Repeat for x = 5

y = 5*x - 14

y = 5*5 - 14

y = 25 - 14

y = 11

We have 11 as the second answer.

--------------

Repeat for x = 6

y = 5*x - 14

y = 5*6 - 14

y = 30 - 14

y = 16

The third answer is 16.

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Repeat for x = 10

y = 5*x - 14

y = 5*10 - 14

y = 50 - 14

y = 36

The last answer is 36.

3 0

3 years ago