David is making his own strawberry yogurt. In his mixture, the number of strawberries is proportional to the amount of milk in cups david uses 2 1/2cups of milk for every 14 strawberries. If david uses 203 strawberries, how much milk will he need?

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.

Answer:

simplified answer: its letter c

Step-by-step explanation:

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 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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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

The gravitational potential energy of a 3 kg toy car resting at the top of a 0.45-meter ramp is calculated using the formula PE = mgh, giving us a result of 13.2385 Joules.

The calculation of gravitational potential energy for a 3 kg toy car at the top of a 0.45-meter ramp. Gravitational potential energy (PE) is calculated using the formula PE = mgh, where m stands for mass, g is the acceleration due to gravity (usually 9.81 m/s² on Earth's surface), and h is the height above the reference point. In this case, PE = 3 kg × 9.81 m/s² × 0.45 m, which gives us a gravitational potential energy of 13.2385 Joules. As the car is at rest on the top of the ramp and has not yet moved, its potential energy is at its maximum at that point.

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

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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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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