You are saving money to buy a new game system you received $50 as a birthday gift from your grandparents. You want to save $25 a week from mowing lawns.

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:

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 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:

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

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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Answer:37

Step-by-step explanation:

12•12=144

35•35=1225

1225+144=1369

Square root 1369=37

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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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

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]

The shaded region is by assumption the region which is not covered by the semicircle in in given rectangles.

The area of the shaded region is given by 15.48 sq. ft.

A semicircle is a circle cut in half. Thus, one circle produces two semicircle.

Firstly we will find the area of the rectangle and then subtract the area of the semicircle to find the are of the shaded region.

Since the radius of the semicircle is equal to width of the rectangle(6 ft), thus the length of the diameter of the circle( twice the radius which is 12 ft) serves as length of the considered rectangle.

Thus, we have:

[tex]\text{Area of the given rectangle\:} = 6 \times 12 = 72 \: \rm ft^2[/tex]

Since the semicircle is having radius of 6 ft, thus:

[tex]\text{Area of semicircle} = \dfrac{\pi r^2}{2} = \dfrac{3.14 \times 6^2}{2} = 56.52 \: \rm ft^2[/tex]

Thus, area of the shaded region will be equal to area of rectangle - area of semicircle = 72 - 56.52 = 15.48 sq. ft.

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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

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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Answer:

6 to the tenth power over 7 to the sixth power

Step-by-step explanation:

Given phrase,

6 to the fifth power over 7 cubed all raised to the second power,

[tex]\implies (\frac{6^5}{7^3})^2[/tex]

By using [tex](a^m)^n=a^{mn}[/tex]

[tex]=\frac{6^{5\times 2}}{7^{3\times 2}}[/tex]

[tex]=\frac{6^{10}}{7^6}[/tex]

= 6 to the tenth power over 7 to the sixth power

Simplify the expressions

(6⁵/7³)² = 2143588816/117649

(6⁷/7¹⁰) = 279936/282475249

(6¹⁰/7⁶) = 60466176/117649

(6³/7) = 216/7

(12⁵/14³) = 90855/1001

To simplify the given expressions, we can calculate the numerical values and perform the necessary operations. Let's evaluate each expression:

(6⁵/7³)²:

First, calculate the numerator and denominator:

Numerator: 6⁵ = 6 × 6 × 6 × 6 × 6 = 7776

Denominator: 7³ = 7 × 7 × 7 = 343

Now, substitute the values into the expression and square the result:

(7776/343)² = (7776/343) × (7776/343) = 2143588816/117649

The simplified form is 2143588816/117649.

(6⁷/7¹⁰):

Calculate the numerator and denominator:

Numerator: 6⁷ = 6 × 6 × 6 × 6 × 6 × 6 × 6 = 279936

Denominator: 7¹⁰ = 7 × 7 × 7 × 7 × 7 × 7 × 7 × 7 × 7 × 7 = 282475249

Substitute the values into the expression:

279936/282475249

This expression cannot be simplified further.

(6¹⁰/7⁶):

Calculate the numerator and denominator:

Numerator: 6¹⁰ = 6 × 6 × 6 × 6 × 6 × 6 × 6 × 6 × 6 × 6 = 60466176

Denominator: 7⁶ = 7 × 7 × 7 × 7 × 7 × 7 = 117649

Substitute the values into the expression:

60466176/117649

This expression cannot be simplified further.

(6³/7):

Calculate the numerator and denominator:

Numerator: 6³ = 6 × 6 × 6 = 216

Denominator: 7

Substitute the values into the expression:

216/7

This expression cannot be simplified further.

(12⁵/14³):

Calculate the numerator and denominator:

Numerator: 12⁵ = 12 × 12 × 12 × 12 × 12 = 248832

Denominator: 14³ = 14 × 14 × 14 = 2744

Substitute the values into the expression:

248832/2744 = 90855/1001

The simplified form is 90855/1001.

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Simplify 6 to the fifth power over 7 cubed all raised to the second power. 6 to the seventh power over 7 to the tenth power 6 to the tenth power over 7 to the sixth power 6 cubed over 7 12 to the fifth power over 14 cubed

(6⁵/7³)²

(6⁷/7¹⁰)

(6¹⁰/7⁶)

(6³/7)

(12⁵/14³)