If f(x) = –x3, g(x) = 3x2 – 1, and h(x) = 2x + 5, what is the degree of [f o g o h](x)?

a-2
b-3
c-5
d-6

The roll of plain wrapping paper sold is  25.

The  roll of shiny wrapping paper sold is 30.

a + b = 55 equation 1

14a + 20b = 950 equation 2

Where:

a =  roll of plain wrapping paper sold

b = roll of shiny wrapping paper sold

Multiply equation 1 by 14

14a + 14b = 770 equation 3

Subtreact equation 3 from equation 2

6b = 180

Divide both sides by 6

b  = 30

Subtract 30 from 55

55 - 33 = 25

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I'm assuming that the 1/3 is an exponent.

If so, then  

Which is the cube root of 2. Raising any value to the 1/3 power is the same as taking the cube root.

Answer:

If you wanted to predict the value of the y variable when the x variable is 15, you would be extrapolating the data.

Step-by-step explanation:

If you wanted to predict the value of the y variable when the x variable is 15, you would be Extrapolating the data.

Extrapolation means, estimating the value of a variable beyond its given range.

Number of chicken = 26

And, Number of dogs = 17

Mathematical expression is defined as the collection of the numbers variables and functions by using operations like addition, subtraction, multiplication, and division.

Given that;

A farmer has a collection of chickens and dogs.

And, all together there are 120 legs and 43 heads.

Hence, Number of chicken = x

And, Number of dogs = y

Thus, We get;

⇒ x + y = 43  

⇒ x = 43 - y .. (i)

⇒ 2x + 4y = 120

⇒ x + 2y = 60 .. (ii)

Substitute value of x from (i) to (ii);

⇒ x + 2y = 60

⇒ 43 - y + 2y = 60

⇒ y = 60 - 43

⇒ y = 17

And, From (i);

⇒ x = 43 - y

⇒ x = 43 - 17

⇒ x = 26

Thus,  Number of chicken = 26

And, Number of dogs = 17

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[tex]c=14[/tex]

There is no extraneous solution because as we put the value of c in equation (1), it does not make the equation (1) not define.

Step-by-step explanation:

Given :

[tex]\dfrac {c-4}{c-2} = \dfrac {c-2}{c+2} - \dfrac{1}{2-c}[/tex]                   --------- (1)

Calculation :

[tex]\dfrac {c-4}{c-2} = \dfrac {(c-2)^2+(c+2)}{(c+2)(c-2)}[/tex]

[tex]c-4 = \dfrac {(c^2-4c+4)+(c+2)}{(c+2)}[/tex]

[tex](c-4)(c+2) = (c^2-3c+6)[/tex]

[tex](c^2-2c-8) = (c^2-3c+6)[/tex]

[tex]c=14[/tex]

There is no extraneous solution because as we put the value of c in equation (1), it does not make the equation (1) not define.

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Answer:

14 and Zero

Step-by-step explanation:

The expression that can be simplified by combining like terms is 16 c^2 + 8cd + 7c - 8d, where 8cd and -3dc are like terms and can be combined to simplify the expression to 16c^2 + 5cd + 7c - 8d.

The expression 16 c2 + 8cd + 7c - 8d can be simplified by combining like terms. Like terms in an algebraic expression are terms that have the same variable raised to the same power. Here, the terms 8cd and -3dc are like terms and can be combined.

Here's the step-by-step simplification:

This process of combining like terms helps in simplifying algebraic expressions, making them easier to work with.

After combining like terms, it's important to always check the answer to see if the simplification is reasonable and if there are any other like terms that can be combined.

So, the simplified expressions are:

1. [tex]\(6d^2 + (5c - 3d)d + 8\)[/tex]

2. [tex]\(12c^2 - (c(8d - 3)) + 4\)[/tex]

3. [tex]\(7d^2 - 3c^2 + 6d - c\)[/tex]

4. [tex]\(16c^2 + c(8d + 7) - 8d\)[/tex]

Let's simplify each expression by combining like terms:

1. [tex]\(6d^2 + 5cd - 3dc + 8\)[/tex]

  Combine like terms:

  [tex]\(6d^2 + (5cd - 3dc) + 8\)[/tex]

 [tex]\(6d^2 + (5c - 3d)d + 8\)[/tex]

2. [tex]\(12c^2 - 8d^2c + 3dc + 4\)[/tex]

  Combine like terms:

[tex]\(12c^2 - (8d^2c - 3dc) + 4\)[/tex]

  [tex]\(12c^2 - (c(8d - 3)) + 4\)[/tex]

3. [tex]\(7d^2 - 3c^2 + 6d - c\)[/tex]

  No like terms to combine.

4. [tex]\(16c^2 + 8cd + 7c - 8d\)[/tex]

  Combine like terms:

  [tex]\(16c^2 + (8cd + 7c) - 8d\)[/tex]

  [tex]\(16c^2 + c(8d + 7) - 8d\)[/tex]

Answer:

7 years

Step-by-step explanation:

An initial investment of $200 is now valued at $350.

The annual interest rate is 8% compounded continuously.

Formula:

[tex]A=Pe^{rt}[/tex]

Where,

A is amount, A=350

P is principle, P=200

R is rate of interest, r=0.08

t is time, t=?

Substitute the value into formula

[tex]350=200e^{0.08\cdot t}[/tex]

[tex]e^{0.8t}=1.75[/tex]

Taking ln both sides

[tex]0.08t = ln(1.75)[/tex]

[tex]t=6.99\approx 7[/tex]

Hence, 7 years ago money was invested.

Answer:

Its 7 years

Step-by-step explanation:

Rene will save $0.94 if she uses her coupon and buys the name brand cookies instead of the store brand cookies.

To find out how much Rene will save if she uses her coupon and buys the name brand cookies instead of the store brand cookies, we have to first determine how much the name brand cookies will cost after applying the coupon. We subtract the value of the coupon i.e. $3.25 from the original cost of name brand cookies which is $7.89. The result comes out to be $4.64.

To calculate the savings, we now have to subtract the cost of name brand cookies after coupon ($4.64) from the original cost of the store brand cookies ($5.58). Hence, the savings amount to $5.58 - $4.64 which equals $0.94

So, if Rene uses her coupon and buys the name brand cookies instead of the store brand cookies, she will save $0.94.

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Answer:

11.18

Step-by-step explanation:

we know that

A median of a triangle is a line segment joining a vertex to the midpoint of the opposing side, bisecting it.

In this problem the line DB is a median of triangle ADC

The line DB is also the altitude of a triangle ADC, because is perpendicular to the side AC

so

AB=BC

we have that

[tex]BC=9\ units[/tex]

therefore

[tex]AB=9\ units[/tex]

the answer is the option

[tex]9[/tex]

Final answer:

The amount of iodine-123 remaining after t hours can be expressed as I(t) = 52 · 0.945^t, where the constant 0.945 is calculated using the half-life of 13 hours.

Explanation:

The half-life of a radioactive isotope is the time it takes for half of the radioactive atoms to decay. For iodine-123, which has a half-life of approximately 13 hours, we can use an exponential decay model to express the amount of iodine-123 remaining after t hours.

The decay formula is generally given by I(t) = I0e-kt, where I0 is the initial amount, k is the decay constant, and t is time. However, the question asks us to express the equation in the form of I(t) = a · bt, which is another common representation for exponential decay.

To find the value of b, we know that after one half-life, b13 = 1/2. Thus, b = (1/2)1/13.
Let's calculate b: b = (0.5)1/13 ≈ 0.945 (rounded to three decimal places).

Since we start with 52 grams of iodine-123, our initial amount (a) is 52. Therefore, the equation for the amount of iodine-123 remaining after t hours is:

I(t) = 52 · 0.945t

The equation to calculate the amount of iodine-123 remaining after t hours is I(t) = a * b^t, where a is the initial amount of iodine-123, b is the decay constant, and t is the time in hours. The decay constant can be calculated using the formula b = 0.693 / t1/2, where t1/2 is the half-life of iodine-123. In this case, the half-life is 13 hours.

The equation that gives the amount of iodine-123 remaining after t hours can be written as:

I(t) = a * b^t

Where:

To calculate the decay constant b, we can use the formula:

b = 0.693 / t1/2

where t1/2 is the half-life of iodine-123. In this case, the half-life is 13 hours, so:

b = 0.693 / 13 = 0.053

Therefore, the equation becomes:

I(t) = a * 0.053^t

Round the value of b to three decimal places, so b = 0.053.

y = k/x

Step-by-step explanation:

Answer:

A) [tex]24\text{ m}^3[/tex].

Step-by-step explanation:

We have been given an image of a square pyramid and we are asked to find the volume of our given pyramid.

[tex]\text{Volume of square pyramid}=\frac{a^2*h}{3}[/tex], where,

a = Base length of square,

h = height of pyramid.

Upon substituting our given values in above formula we will get,

[tex]\text{Volume of square pyramid}=\frac{\text{(3 m)}^2*\text{ 8 m}}{3}[/tex]

[tex]\text{Volume of square pyramid}=\frac{9\text{ m}^2*\text{ 8 m}}{3}[/tex]

[tex]\text{Volume of square pyramid}=\frac{72\text{ m}^3}{3}[/tex]

[tex]\text{Volume of square pyramid}=24\text{ m}^3[/tex]

Therefore, the volume of our given square pyramid is 24 cubic meters and option A is the correct choice.

Answer:

A. 24 m3.

Step-by-step explanation:

The volume of the pencil is approximately 4.05 cm³.

To determine the volume of the pencil, we need to calculate the volumes of the cylindrical body and the conical tip separately and then add them together.

Volume of the cylindrical body:

The formula for the volume of a cylinder is [tex]V = \pi r^2h[/tex], where r is the radius and h is the height.

[tex]V_{cylinder} = \pi (0.5 cm)^2 * 15 cm[/tex]

Volume of the conical tip:

The formula for the volume of a cone is [tex]V = (1/3)\pi r^2h[/tex], where r is the radius and h is the height.

[tex]V_{cone} = (1/3) * \pi (0.5 cm)^2 * 2 cm[/tex]

Now, we can calculate the volumes:

[tex]V_{cylinder} = \pi * (0.5 cm)^2 * 15 cm = 3.53 cm^3\\\\V_{cone} = (1/3) * \pi * (0.5 cm)^2 * 2 cm = 0.52 cm^3[/tex]

Finally, we can add the volumes of the cylindrical body and the conical tip to get the total volume of the pencil:

[tex]Total\ volume = V_{cylinder} + V_{cone} = 3.53 cm^3 + 0.52 cm^3 = 4.05 cm^3[/tex]

Therefore, the volume of the pencil is approximately 4.05 cubic centimeters ([tex]cm^3[/tex]).

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General Idea:

There are few exponent rules that need to be used to simplify the rational exponent to get to the answer.

Exponent rules:

[tex] Rule\; 1: a \cdot a \cdot a \cdot a =a^4\\ \\ Rule\; 2: (a^m)^n=a^{m \times n} [/tex]

Applying the concept:

Rewriting 81 as a number with an exponent using prime factorization.

[tex] 81=3 \times 3 \times 3 \times 3 = 3^4\\ \\ Applying \; Rule \; 1, \; we\; get\; \\\\81^{\frac{3}{4}}=(3^4)^{\frac{3}{4}} \\ \\ Applying \; Rule \; 2,\; we\; get\\ \\ 3^{4 \times\frac{3}{4} } = 3^{\frac{12}{4} }=3^3=3 \times 3 \times 3 = 27 [/tex]

Conclusion:

The above explanation explains why [tex] 81^{\frac{3}{4}} =27 [/tex]