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2 years ago

Write a quadratic function f whose only zero is 7

1 answer:

5 0

well, we know is a quadratic, namely of degree 2, so it has two solutions and namely two factors, and whose only zero is 7, well, we can do that by giving that only zero a multiplicity of 2, so we get two of the same zero.

$\begin{cases} x=7\implies &x-7=0\\ x=7\implies &x-7=0 \end{cases}\qquad \implies \qquad \begin{array}{llll} \stackrel{\textit{multiplicity of 2}}{(x-7)(x-7)}=\stackrel{0}{y} \\\\\\ x^2-14x+49=y \end{array}$

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Solve the equation 3x-4=8-x

solmaris [256]

Answer:

x=3

Step-by-step explanation:

3x-4=8-x(Group like terms)

3x+x=8+4(Add both sides)

4x=12(After adding you will proceed to divide both sides by 4)

x=3(x is 3 because 4 can divide 12 3 times that's why we have x as equal to 3)

5 0

3 years ago

Find the period of the graph

olya-2409 [2.1K]

Answer:

240 degrees

Step-by-step explanation:

Looking at the graph we can see that the first peak is at -30 degrees and the second peak is at 210 degrees.

210 - (-30) = 210 + 30 = 240

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3 years ago

Which statements about residuals are true for the least-squares regression line? I. A random scattering of residuals indicates t

Daniel [21]

Answer:

II. The sum of the residuals is always 0.

Step-by-step explanation:

A least squares regression line is a standard technique in regression analysis used to make the vertical distance obtained from the data points running to the regression line to become very minimal or as small as possible.

For any least-squares regression line, the sum of the residuals is always zero.

Basically, residuals are used to measure or determine whether or not the line of regression is a good fit or match for the data by subtracting the difference between them i.e the predicted y value and the actual y value, for the x value respectively.

Hence, the statement about residuals which is true for the least-squares regression line is that the sum of the residuals is always zero (0).

7 0

3 years ago

How can i differentiate this equation?

Dmitry_Shevchenko [17]

$\bf y=\cfrac{2x^2-10x}{\sqrt{x}}\implies y=\cfrac{2x^2-10x}{x^{\frac{1}{2}}} \\\\\\ \cfrac{dy}{dx}=\stackrel{\textit{quotient rule}}{\cfrac{(4x-10)(\sqrt{x})~~-~~(2x^2-10x)\left( \frac{1}{2}x^{-\frac{1}{2}} \right)}{\left( x^{\frac{1}{2}} \right)^2}} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~(2x^2-10x)\left( \frac{1}{2\sqrt{x}} \right)}{\left( x^{\frac{1}{2}} \right)^2} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~\left( \frac{2x^2-10x}{2\sqrt{x}} \right)}{x}$

$\bf\cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~\left( \frac{2x^2-10x}{2\sqrt{x}} \right)}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{ \frac{(4x-10)(\sqrt{x})(2\sqrt{x})~~-~~(2x^2-10x)}{2\sqrt{x}}}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})(2\sqrt{x})~~-~~(2x^2-10x)}{2x\sqrt{x}}$

$\bf \cfrac{dy}{dx}=\cfrac{(4x-10)2x~~-~~(2x^2-10x)}{2x\sqrt{x}}\implies \cfrac{dy}{dx}=\cfrac{8x^2-20x~~-~~(2x^2-10x)}{2x\sqrt{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{8x^2-20x~~-~~2x^2+10x}{2x\sqrt{x}} \implies \cfrac{dy}{dx}=\cfrac{6x^2-10x}{2x\sqrt{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{2x(3x-5)}{2x\sqrt{x}}\implies \cfrac{dy}{dx}=\cfrac{3x-5}{\sqrt{x}}$

8 0

3 years ago

What is the slope of the line that passes throught the points (1,1) and (-1,-5)

Amiraneli [1.4K]

$m = \dfrac{ \textrm{rise}}{\textrm{run}} = \dfrac{1 - -5 }{1 - -1} = \dfrac{6}{2} = 3$

Slope is 3

6 0

3 years ago