Generally, how would one work out the surface area of a frustum?

Answer:

[tex]A) 7[/tex]

Step-by-step explanation:

[tex]-2x+4=2\left(4x-3\right)-3\left(-8+4x\right)[/tex]

[tex]2\left(4x-3\right)-3\left(-8+4x\right)[/tex]

[tex]2 (4x - 3)[/tex]

[tex]\longrightarrow8x-6[/tex]

[tex]\longrightarrow 8x-6-3\left(-8+4x\right)[/tex]

[tex]-3 (-8 + 4x)[/tex]

[tex]\longrightarrow 24-12x[/tex]

[tex]8x-6+24-12x[/tex]

Combine like terms:

[tex]8x-12x-6+24[/tex]

Add: [tex]8x-12x=-4x[/tex]

Add: [tex]-6+24=18[/tex]

[tex]\longrightarrow-2x+4=-4x+18[/tex]

Subtract 4 from both sides:

[tex]\longrightarrow-2x+4-4=-4x+18-4[/tex]

[tex]\longrightarrow-2x=-4x+14[/tex]

Add 4x to both sides:

[tex]\longrightarrow-2x+4x=-4x+14+4x[/tex]

[tex]\longrightarrow2x=14[/tex]

Divide both sides by 2:

[tex]\longrightarrow\frac{2x}{2}=\frac{14}{2}[/tex]

[tex]\longrightarrow x=7[/tex]

____________________________

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The derivative of f(x) at x = -1 is -8.

Derivative is the rate of change of a function with respect to a variable. Derivatives are fundamental to the solution of problems in calculus and differential equations.

According to the given problem,

f(x) = [tex]\frac{8}{x}[/tex]

Derivative of f(x),

⇒ f'(x) = [tex]-\frac{8}{x^{2} }[/tex]

At x = -1,

⇒ f'(-1) =  [tex]-\frac{8}{(-1)^{2} }[/tex]

          = -8

Hence, we can conclude, the derivative of f(x) at x = -1 is -8.

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Answer:

True

Step-by-step explanation:

Given statement : The base of an exponential function cannot be a negative number.

We need to check whether the given statement is true or not.

The general form of an exponential function is

[tex]f(x)=ab^x[/tex]

where, a is the initial value and b is the base of exponential function.

The value of base b is always greater than 0 because

1. The terms like [tex](-3)^\frac{1}{2}[/tex] is an imaginary number and it make no sense. So, the base of an exponential function cannot be a negative number.

2. It b=0, then [tex](0)^x=0[/tex], which is a constant function. So, the base of an exponential function cannot be 0.

It means b>0.

Therefore, the given the given statement is true.

Final answer:

The base of an exponential function must be positive, as a negative base can lead to undefined values when raised to non-integer exponents. The rate of growth of an exponential function with a positive base near one can be approximated by 1+x for small x. Bases for exponential functions can often be connected to Euler's number e (approximately 2.71818).

Explanation:

The statement that the base of an exponential function cannot be a negative number is true. The base of an exponential function needs to be positive because a negative base can lead to undefined or complex values when raised to a fractional or irrational exponent. For example, the exponential function with base e (where e is the Euler's number, approximately equal to 2.71818) demonstrates natural growth, and its rate of growth for small x values approximates to 1+x. This approximation indicates how for a small change in x, the change in the value of the function is nearly proportional when the base is a positive number close to one.

Considering other bases, they can be related to base e using the identity ax = eln(a)x, where a is a positive real number, and ln(a) is the natural logarithm of a. Thus, whether for common or natural bases, the exponential function is well-defined only for positive bases.

The third derivative of the function is [tex]y^{'''}=\dfrac{\left(3y^2+12xy-x\right)y''+y'\left(6yy'+12xy'+12y-1\right)-\left(6y-1\right)y'-6yy'}{\left(3y^2+x\right)^2}-\dfrac{2\left(6yy'+1\right)\left(\left(3y^2+12xy-x\right)y'-y\cdot\left(6y-1\right)\right)}{\left(3y^2+x\right)^3}[/tex]and the value at the point x = 1 is 42

How to determine the third derivative at the point x = 1

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

[tex]x^2 + xy + y^3 = 1[/tex]

Differentiate implicitly

So, we have

[tex]3y^2y' + xy'+y+2x=0[/tex]

Make y' the subject of formula

So, we get

[tex]y'=-\dfrac{y+2x}{3y^2+x}[/tex]

Differentiate the second time

Using a graphing tool, we have

[tex]y''=\dfrac{\left(3y^2+12xy-x\right)y'-6y^2+y}{\left(3y^2+x\right)^2}[/tex]

Differentiate the third time to get the third derivative

Using a graphing tool, we have

[tex]y^{'''}=\dfrac{\left(3y^2+12xy-x\right)y''+y'\left(6yy'+12xy'+12y-1\right)-\left(6y-1\right)y'-6yy'}{\left(3y^2+x\right)^2}-\dfrac{2\left(6yy'+1\right)\left(\left(3y^2+12xy-x\right)y'-y\cdot\left(6y-1\right)\right)}{\left(3y^2+x\right)^3}[/tex]

Recall that

x = 1

Calculating y, we have

[tex]1^2 + (1)y + y^3 = 1[/tex]

[tex]1 + y + y^3 = 1[/tex]

[tex]y^3 + y = 0[/tex]

Factorize

[tex]y(y^2 + 1) = 0[/tex]

So, we have

y = 0 or [tex]y^2 + 1 = 0[/tex]

The equation [tex]y^2 + 1 = 0[/tex] will give a complex solution

So, we have

x = 1 and y = 0

Calculating y', we have

[tex]y'=-\dfrac{0+2(1)}{3 * 0^2+1}[/tex]

[tex]y'=-\dfrac{2}{1}[/tex]

y' = -2

Calculating y", we have

[tex]y''=\dfrac{\left(3y^2+12y-1\right)y'-6y^2+y}{\left(3y^2+1\right)^2}[/tex]

[tex]y''=\dfrac{\left(3(0)^2+12(1)(0)-1\right)(-2)-6(0)^2+0}{\left(3(0)^2+1\right)^2}[/tex]

[tex]y''=\dfrac{\left2}{1}[/tex]

y" = 2

Calculating y", we have

[tex]y^{'''}=\dfrac{\left(3y^2+12xy-x\right)y''+y'\left(6yy'+12xy'+12y-1\right)-\left(6y-1\right)y'-6yy'}{\left(3y^2+x\right)^2}-\dfrac{2\left(6yy'+1\right)\left(\left(3y^2+12xy-x\right)y'-y\cdot\left(6y-1\right)\right)}{\left(3y^2+x\right)^3}[/tex]

Simplifying the denominators, we have

[tex](3y^2 + x)^2 = (3(0)^2 + 1)^2 = 1[/tex]

[tex](3y^2 + x)^3 = (3(0)^2 + 1)^3 = 1[/tex]

So, we have

[tex]y^{'''}=\dfrac{\left(3y^2+12xy-x\right)y''+y'\left(6yy'+12xy'+12y-1\right)-\left(6y-1\right)y'-6yy'}{1}-\dfrac{2\left(6yy'+1\right)\left(\left(3y^2+12xy-x\right)y'-y\cdot\left(6y-1\right)\right)}{1}[/tex]

Divide

[tex]y^{'''}=[\left(3y^2+12xy-x\right)y''+y'\left(6yy'+12xy'+12y-1\right)-\left(6y-1\right)y'-6yy']-[2\left(6yy'+1\right)\left(\left(3y^2+12xy-x\right)y'-y\cdot\left(6y-1\right)\right)][/tex]

Simplifying each term:

[tex](3y^2+12xy-x)y''+y'(6yy'+12xy'+12y-1)-(6y-1)y'-6yy' = (3(0)^2+12(1)(0)-(1))(2) + (-2)(6(0)(-2) +12(1)(-2) + 12(0) - 1) - (6(0) - 1)(-2) - 6(0)(-2)[/tex]

[tex](3y^2+12xy-x)y''+y'(6yy'+12xy'+12y-1)-(6y-1)y'-6yy' = 46[/tex]

Also, we have

[tex]2(6yy'+1)((3y^2+12xy-x)y'-y(6y-1)) = 2(6(0)(-2) + 1)((3(0)^2 + 12(1)(0) - 1)(-2) - 0(6(0)-1))[/tex]

[tex]2(6yy'+1)((3y^2+12xy-x)y'-y(6y-1)) = 4[/tex]

So, the expression becomes (by substitution)

[tex]y^{'''}= 46 -4[/tex]

This gives

[tex]y^{'''}= 42[/tex]

Hence, the third derivative at the point x = 1 is 42

Answer:

1 + 1 = 2

2 - 1 = 1

The line's slopes are cute

Step-by-step explanation:

I hope your lines enjoy their equations, you are a great parent to your lines. =D

The vertical asymptotes of the function y = 2cot(3x) + 4 are  A.x = π/3  C. x = 2π D.x = 0

The function is given as:

y = 2cot(3x) + 4

The above function is a cotangent function, represented as:

y = Acot(Bx +C) + D

By comparison, we have:

B = 3

The vertical asymptotes are then calculated using:

[tex]x = \frac{\pi}{B}n[/tex], where n are integers

Substitute 3 for B

[tex]x = \frac{\pi}{3}n[/tex]

Using the above format, the vertical asymptotes in the options are  A.x = π/3  C. x = 2π D.x = 0

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The rate of change of radius of the circle when the radius is 10 cm is

dr / dt = 1/2π cm / sec

What is implicit differentiation?

In mathematics, differentiation is the process of determining the derivative, or rate of change, of a function. Differentiation is a technique for determining a function's derivative. Differentiation is a mathematical procedure that determines the instantaneous rate of change of a function based on one of its variables. The process of finding the derivative of dependent variable in an implicit function by differentiating each term separately

Given data ,

Oil is poured on a flat surface and it spreads out forming a circle

The area of the circle is increasing at a rate of 10 cm²/sec

Let the area of the circle be = A

So , the equation is dA² / dt² =  10 cm²/sec

The area of the circle = πr²

So , dA² / dt² = 2π dr / dt

2πr dr / dt = 10 cm²/sec

Substituting the value of r in the equation , we get

2π x 10 dr / dt = 10 cm²/sec

20π dr / dt = 10 cm²/sec

Divide both sides of the equation be 20π

dr / dt = ( 1/2π ) cm / sec

Therefore , the rate of change of radius of the circle is ( 1/2π ) cm / sec

Hence , The rate of change of radius of the circle when the radius is 10 cm is dr / dt = 1/2π cm / sec

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Answer-

The inequality that models this situation is,

[tex]\boxed{\boxed{B.\ 6x+3\leq 27}}[/tex]

Solution-

The total money Charlie has = $27

He wants to order 6 sandwiches and a kid's meal of $3

Let us assume the price of each sandwiches is x.

So, the amount he will be spending for sandwiches is $6x

The total amount he will be spending is $(6x+3)

As he has only $27, so he can spend any amount less than or equal to 27

So the inequality becomes,

[tex]\Rightarrow 6x+3\leq 27[/tex]

Therefore, the inequality that models this situation is [tex]6x+3\leq 27[/tex]

If we equate the two given equations y = m^4 and y = n^3, we can express mn as m^(7/3) or as n^(7/4). This is done by taking the cube root and square root on both sides of the equated equation to isolate m and n.

If we have two equations, y = m^4 and y = n^3, where y > 1, we can equate these two equations since they are both equal to same value 'y'. So the equation will be m^4 = n^3. To find the value of mn, we want to express it in terms of 'y'. So we take cube root on both sides, it will be m = (n^3)^(1/4) = n^(3/4). And square root on both sides will give n = (m^4)^(1/3) = m^(4/3). Now mn = m * m^(4/3) = m^(1 + 4/3) = m^(7/3). Similarly, mn = n * n^(3/4) = n^(1 + 3/4) = n^(7/4). Hence, mn can be expressed either as m^(7/3) or n^(7/4).

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Answer:

simplified answer: its letter c

Step-by-step explanation:

area of yard = 41 x 60 = 2460 square feet

walk is 4 feet wide

 41-4 =37, 60-4 = 56

37*56 = 2072

2460-2072 = 388

 area of walk is 388 square feet