Anthony's lovker is (x+4) ft . long qnd (x+2) ft wide

Write and equation for the shift of the parent graph y=1/x

SashulF [63]

You can think of this equation as $\bf f(x)=\cfrac{1}{x}\implies f(x)=(x)^{-1}$

and thus apply the transformations to it as such

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\\\\
--------------------\\\\$

$\bf \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\$

$\bf \bullet \textit{ vertical shift by }{{ D}}\\
\left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\
\left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}$

1) D = 2, C = +3

2) D = -12, C = -2

3) C = +6, D = 3

4) C = -7, D = -7

7 0

3 years ago

Satoranaing

Natali [406]

The statements that are true about the given points on the line are; Options B, C and D

<h3>How to identify points on a Plane?</h3>

A plane is defined by a line and a point outside of it. Meanwhile, a line is defined by two points. Thus, whenever we have 3 non-collinear points, we can define a plane.

Let us check each of the given options;

A) There are exactly two planes that contain points A, B, and F;

If these points are collinear, they can't make a plane and If they are not collinear, they define a plane.

Since we can't make two planes with them, then the statement is false.

B) There is exactly one plane that contains points E, F, and B;

Due to the same reasoning in part A, the statement is true assuming the points are not collinear.

C) The line that can be drawn through points C and G would lie in plane X;

The statement is true because both points C and G lie on plane X, and as such the line that connects them also should lie on the same plane.

D) The line that can be drawn through points E and F would lie in plane Y; Similar to above, this statement is true.

E) The only points that can lie on plane Y are points E and F.

This statement is false because infinite points can lie on a plane.

Read more about plane points at; brainly.com/question/11958640

#SPJ1

4 0

2 years ago

Simplify the expression. Explain each step 2 + (5 + y)

Inessa [10]

Answer:

7y

Step-by-step explanation:

2+(5+y)

2+5y

=7y.............

6 0

3 years ago

Write and solve an inequality for the possible values of x.

tatuchka [14]

The possible values of x are x > 0.5

The given diagram is a quadrilateral. From the diagram we can see that;

3x + 2 > x + 3 (according to the length)

Solve the resulting inequality

3x - x > 3 - 2

2x > 1

x > 1/2

x> 0.5

Hence the possible values of x are x > 0.5

Learn more on inequality here: brainly.com/question/11613554

4 0

2 years ago

Please help <br> I need to turn it in now

morpeh [17]

Answer:

Step-by-step explanation:

x=rcosθ=4 cos (-40)=4 cos 40≈3.064

y=r sin θ=4 sin (-40)=-4 sin 40≈-2.571

the vector is (3.064,-2.571)

8 0

3 years ago