The function table below is intended to represent the relationship y=-5x+1. However, one of the entries for y does not correctly fit the relationship with x.
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Answer:
Step-by-step explanation:
none of the answers are correct
Related Questions
To make an open box from a 175cm by 100cm piece of cardboard, equal-sized squares will be cut from each of the four corners and then the sides will be folded up. What is the approximate volume of the largest possible box that can be made?Group of answer choices:A) 324,146 cm^3B)162,073cm^3C) 251,707cm^3D)189,640cm^3
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Given that dimensions of the piece of cardboard are:
[tex]\begin{gathered} l=175\text{ }cm \\ w=100\text{ }cm \end{gathered}[/tex]Where "l" is the length and "w" is the width, you can determine that it has the shape of a rectangle.
You know that equal-sized squares will be cut from each of the four corners and then the sides will be folded up. Then, you can make the following drawing:
By definition, the volume of a rectangle is:
[tex]Volume=length\cdot width\cdot height[/tex]In this case, you can set up that:
[tex]\begin{gathered} length=175-2x \\ width=100-2x \\ height=x \end{gathered}[/tex]Therefore, you can write this equation:
[tex]V=(175-2x)(100-2x)(x)[/tex]Expand it:
[tex]V=(175-2x)(100x-2x^2)[/tex][tex]V=(175)(100x)-(175)(2x)-(2x)(100x+(2x)(2x^2)[/tex][tex]V=4x^3-550x^2+17500x[/tex]Now you need to derivate it using the Power Derivative Rule:
[tex]\frac{d}{dx}(x^n)=nx^{n-1}[/tex]Then:
[tex]V^{\prime}=(4)(3)x^2-(550)(2)x+17500[/tex][tex]V^{\prime}=12x^2-1100x+17500[/tex]Make the equation equal to zero and sove for "x":
[tex]12x^2-1100x+17500=0[/tex]Use the Quadratic Formula:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]Substituting:
[tex]\begin{gathered} a=12 \\ b=-1100 \\ c=17500 \end{gathered}[/tex]You get:
[tex]\begin{gathered} x_1\approx20.49 \\ x_2\approx71.18 \end{gathered}[/tex]Therefore, you can make the following Sign Chart:
Then, you can substitute these:
[tex]\begin{gathered} x=20 \\ x=50 \\ x=72 \end{gathered}[/tex]Into the factorize form of the derivated function:
[tex]V=(x+\frac{275-25\sqrt{37}}{6})(x-\frac{275+25\sqrt{37}}{6})[/tex][tex]V=(x+\frac{275-25\sqrt{37}}{6})(x-\frac{275+25\sqrt{37}}{6})[/tex]Solve the equation. Check your solution.20x - 4= 50x + 2x=(Simplify your answer. Type an integer or a simplified fraction.)
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20x - 4 = 50x + 2
Solve for x
20x - 50x =
Convert the polar equation r=3 to a Cartesian equation.x^2+y^2=√3x^2+y^2=3x^2+y^2=9
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For the given equation:
[tex]\begin{gathered} \text{Polar form: } \\ r=3 \\ \\ \text{Cartesian form:} \\ x^2+y^2=3^2 \\ x^2+y^2=9 \end{gathered}[/tex]2x + 4x = 3x + 3x Solve for x.
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You have the following expression:
2x +4x = 3x + 3x
in order to solve for x, proceed as follow:
2x +4x = 3x + 3x simplify like terms both sides
6x = 6x
Due to the previous result is the trivial solution, it means that the equation has infinite solutions.
Evaluate the expression [tex]9 + 7 - 3 \times 3 - 2[/tex]
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Sean, Kevin and Bill take classes at both JJC and CSU. Sean takes 8 credits at JJC and 4 credits at CSU; Kevin takes 10 credits at JJC and 6 at CSU: Bill takes 6 credits at JJC and 4 at CSU; the cost per credit at JJC is $103 and at CSU is $249. a) Write a matrix A that gives the credits each student is taking and B that gives the cost per credit at each school. b) Find the dimension of A and B. c) Find the product AB and the names of its rows and columns.
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ANSWER:
a)
[tex]\begin{gathered} A=\begin{pmatrix}8 & 4 \\ 10 & 6 \\ 6 & 4\end{pmatrix} \\ B=\begin{pmatrix}103 \\ 249\end{pmatrix} \end{gathered}[/tex]b)
Dimension A = 3 x 2
Dimension B = 2 x 1
c)
Cost of credits
Sean $1820
Kevin $2524
Bill $1614
[tex]\begin{pmatrix}Sean \\ \: Kevin \\ \: Bill\end{pmatrix}\begin{pmatrix}1820 \\ \: 2524 \\ \: 1614\end{pmatrix}[/tex]STEP-BY-STEP EXPLANATION:
With the help of the statement, we create the matrices A and B:
[tex]\begin{gathered} A=\begin{pmatrix}8 & 4 \\ 10 & 6 \\ 6 & 4\end{pmatrix}\rightarrow3\times2 \\ B=\begin{pmatrix}103 \\ 249\end{pmatrix}\rightarrow2\times1 \end{gathered}[/tex]Now, we calculate the product just like this:
[tex]\begin{gathered} \text{Product }A\cdot B=\begin{pmatrix}8\cdot103+4\cdot249 \\ 10\cdot103+6\cdot249 \\ 6\cdot103+4\cdot249\end{pmatrix}=\begin{pmatrix}1820 \\ \: 2524 \\ \: 1614\end{pmatrix} \\ \text{Product }A\cdot B=\begin{pmatrix}Sean \\ Kevin \\ Bill\end{pmatrix}\begin{pmatrix}1820 \\ 2524 \\ 1614\end{pmatrix} \end{gathered}[/tex]I need help answering the 2nd part of this question
Answers
An identity function has one output value for each input value.
G is an identity function.
To find its equation apply the slope-intercept form:
y=mx +b
Where:
m= slope ( rise / run)
b= y-intercept ( where the function crosses the y axis)
By looking at the graph we can see that it crosses (0,0) so, b= 0.
And for every unit to right along the x-axis, it goes up by 1 unit along the y axis.
So, m= 1/1 = 1
Equation:
y = x
I need help with review on functions
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The answer is Option 2
Its not a function
The domain and range of the function is shown at the various endpoints of the graph.
At every point you can always identify a value for x (the input value) and also a corresponding value for y (the output value).
Please note the endpoints marked in blue.
These are the endpoints, and for every point, you will have (x, y).
However, note that at the endpoints, the values of y remains the same.
For every relation to be a function, there must be exactly one corresponding y value for every x value. As shown by the blue markings on the graph, the x values on the horizontal axis both have the same y value (which is zero).
Hence, the relationship shown by the graph is not a function
The figure on the right is a scale drawing of the figure on the left. What is the scale factor?
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In order to find the scale factor, we just need to divide one side of the right figure by the corresponding side in the left figure.
So, taking the sides SU and PR, we have:
[tex]\text{scale}=\frac{SU}{PR}=\frac{12}{8}=1.5[/tex]So the scale factor is 1.5.