Yvonne put $4000 in a savings account. At the end of three years, the account had earned $960 in simple interest. A. How much does she have in her own account at the end of three years? B. What annual simple interest rate did the account grow? Show your work. C. How many more dollars would she have in her account if the interest rate were 1% greater? Show your work.

Answer:

[tex]3a^{4}[/tex] .

Step-by-step explanation:

Given : [tex]27a^{12}[/tex]

To Find:The cube root of [tex]27a^{12}[/tex]

Solution:

We are supposed to solve [tex]\sqrt[3]{27a^{12}}[/tex]

So, [tex]\sqrt[3]{27a^{12}}[/tex]

[tex]\Rightarrow (27)^{\frac{1}{3}}(a^{12})^\frac{1}{3}[/tex]

[tex]\Rightarrow (3^3)^{\frac{1}{3}}(a^{4})[/tex]

[tex]\Rightarrow 3a^{4}[/tex]

Hence The cube root of [tex]27a^{12}[/tex] is [tex]3a^{4}[/tex] .

Final answer:

To make the inequality 43 < y - 30 true, add 30 to both sides to get 73 < y. Any number greater than 73, such as 74, 80, or 90, will satisfy the inequality.

Explanation:

To find values that make the inequality 43 < y - 30 true, you simply need to add 30 to both sides of the inequality. This gives you:

73 < y

This means that any value greater than 73 will make the inequality true. Here are three examples:

You can choose any three numbers greater than 73, and they will satisfy the inequality 43 < y - 30.

51 divided by 3 is k, the value of k is 17.

Algebra is the study of abstract symbols, while logic is the manipulation of all those ideas.

The acronym PEMDAS stands for Parenthesis, Exponent, Multiplication, Division, Addition, and Subtraction. This approach is used to answer the problem correctly and completely.

Given;

51 divided by 3 is k

51/3=k

17=k

Therefore, by algebra the answer will be 17

More about the Algebra link is given below.

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Answer: 54

Step-by-step explanation: 43-16=27 times 2 =54

The explanation can be given down in details . Using this information

are the key features needed to graph a polynomial function and to find these key features to sketch a rough graph of a polynomial function.

To graph a polynomial function, you need to identify and understand its key features, which include:

Degree: The highest power of the variable in the polynomial determines its degree. For example, a polynomial of degree 3 would have terms like [tex]ax^{3} +bx^{2} +cx+d.[/tex] The degree influences the behavior of the polynomial, such as whether its ends point up or down.

Leading Coefficient: The coefficient of the term with the highest power of the variable. It influences the end behavior of the polynomial, determining whether the graph rises to the left and rises to the right or falls to the left and falls to the right.

Roots (Zeros): These are the values of [tex]x[/tex] for which the polynomial evaluates to zero. They are the points where the graph crosses the [tex]x[/tex]-axis.

Turning Points: Points where the graph changes direction. These occur at local maxima or minima and can be found by setting the derivative of the polynomial equal to zero and solving for [tex]x[/tex].

To sketch a rough graph of a polynomial function, follow these steps:

Determine the Degree and Leading Coefficient: Identify the highest power of the variable and the coefficient of that term.

Find the Roots (Zeros): Set the polynomial equal to zero and solve for

[tex]x[/tex]. These are the points where the graph intersects the x-axis.

Identify Turning Points: Find the critical points by taking the derivative of the polynomial, setting it equal to zero, and solving for

[tex]x[/tex]. Then, use the second derivative test or analyze the behavior of the function around these points to determine whether they are local maxima or minima.

Plot Key Points: Use the roots and turning points to plot key points on the graph.

Consider End Behavior: Determine whether the leading coefficient is positive or negative. If it's positive, the ends of the graph will point upwards; if it's negative, they will point downwards.

Sketch the Curve: Connect the plotted points smoothly, taking into account the behavior of the polynomial between the key points. Make sure to show any symmetry if present and adjust the curvature based on the behavior around turning points.

Optional: Use a Graphing Tool: If available, utilize graphing software or calculators to verify your sketch and refine it if necessary.

By following these steps, you can sketch a rough graph of a polynomial function and visualize its key features effectively.

COMPLETE QUESTION:

What are the key features needed to graph a polynomial function? Explain how to find these key features to sketch a rough graph of a polynomial function.

Answer: There is 15% error for the erasers

Step-by-step explanation:

Since we have given that

Number of folders he predicted = 25

Number of erasers he predicted = 230

Total number of inventories he predicted is given by

[tex]230+25=255[/tex]

Number of erasers he actually sold = 200

Number of folders he actually sold = 30

Total number of inventories he actually sold is given by

[tex]200+30=230[/tex]

Error in estimation is given by

[tex]230-200=30[/tex]

Percentage of error for the erasers only  

[tex]\frac{30}{200}\times 100\\\\=\frac{30}{2}\\\\=15\%[/tex]

Hence, there is 15% error for the erasers.

Answer:

15 %

Step-by-step explanation:

The amount of a chemical solution is measured to be 2 liters. What is the percent error of the measurement?

25%

The largest rectangular box volume in the first octant, with one vertex on [tex]\(x + 2y + 3z = 9\),[/tex] is [tex]\(\frac{486}{125}\).[/tex]

To find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex on the plane [tex]\(x + 2y + 3z = 9\),[/tex]we can set up the problem using optimization techniques.

Let the coordinates of the vertex of the box that lies on the plane [tex]\(x + 2y + 3z = 9\)[/tex] be[tex]\((x, y, z)\).[/tex] Since the other vertices are on the coordinate planes, the dimensions of the box are [tex]\(x\), \(y\), and \(z\).[/tex]

The volume [tex]\(V\)[/tex]of the rectangular box is given by:

[tex]\[V = x \cdot y \cdot z\][/tex]

Given that this vertex lies on the plane [tex]\(x + 2y + 3z = 9\),[/tex] we have the constraint:

[tex]\[x + 2y + 3z = 9\][/tex]

We need to maximize [tex]\(V\)[/tex] subject to this constraint. To do this, we can use the method of Lagrange multipliers. We introduce a Lagrange multiplier [tex]\(\lambda\)[/tex]  and define the Lagrangian function:

[tex]\[\mathcal{L}(x, y, z, \lambda) = x y z + \lambda (9 - x - 2y - 3z)\][/tex]

To find the critical points, we take the partial derivatives of [tex]\(\mathcal{L}\)[/tex] with respect to [tex]\(x\), \(y\), \(z\), and \(\lambda\)[/tex] and set them to zero:

[tex]\[\frac{\partial \mathcal{L}}{\partial x} = yz - \lambda = 0\]\[\frac{\partial \mathcal{L}}{\partial y} = xz - 2\lambda = 0\]\[\frac{\partial \mathcal{L}}{\partial z} = xy - 3\lambda = 0\]\[\frac{\partial \mathcal{L}}{\partial \lambda} = 9 - x - 2y - 3z = 0\][/tex]

From the first three equations, we can express [tex]\(\lambda\)[/tex] as follows:

[tex]\[\lambda = yz\]\[\lambda = \frac{xz}{2}\]\[\lambda = \frac{xy}{3}\][/tex]

Equating these expressions for [tex]\(\lambda\):[/tex]

[tex]\[yz = \frac{xz}{2} \implies 2yz = xz \implies x = 2y \quad \text{(if \(z \neq 0\))}\]\[yz = \frac{xy}{3} \implies 3yz = xy \implies y = 3z \quad \text{(if \(x \neq 0\))}\][/tex]

Substituting [tex]\(y = 3z\) and \(x = 2y = 2(3z) = 6z\)[/tex] into the constraint [tex]\(x + 2y + 3z = 9\):[/tex]

Now, using [tex]\(z = \frac{3}{5}\):[/tex]

[tex]\[y = 3z = 3 \left(\frac{3}{5}\right) = \frac{9}{5}\]\[x = 6z = 6 \left(\frac{3}{5}\right) = \frac{18}{5}\][/tex]

The dimensions of the box are:

[tex]\[x = \frac{18}{5}, \quad y = \frac{9}{5}, \quad z = \frac{3}{5}\][/tex]

The volume[tex]\(V\)[/tex]  is:

[tex]\[V = x \cdot y \cdot z = \left(\frac{18}{5}\right) \left(\frac{9}{5}\right) \left(\frac{3}{5}\right) = \frac{18 \cdot 9 \cdot 3}{5^3} = \frac{486}{125} = 3.888\][/tex]

Therefore, the volume of the largest rectangular box is:

[tex]\[\boxed{\frac{486}{125}}\][/tex]

The volume of the largest rectangular box with one vertex on the plane x + 2y + 3z = 9 is found using Lagrange multipliers. The maximum volume is 4.5 cubic units. The calculations involve the gradient method and substitution.

To find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the plane x + 2y + 3z = 9, we need to maximize V = xyz subject to the constraint x + 2y + 3z = 9.

We can use the method of Lagrange multipliers for this problem:

This gives us the following system of equations:

From these equations, we can solve for x, y, z, and λ:

Equating and solving, we obtain x = 1.5, y = 1.5, and z = 2.

Finally, substituting these values into V = xyz gives the volume V = (1.5)  imes (1.5)  imes 2 = 4.5.

Answer:

Step-by-step explanation: strike outs per inning pitched