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marta [7]
3 years ago
7

Can this problem be solved using the distance formula?

Mathematics
1 answer:
djyliett [7]3 years ago
5 0
Yes you need the distance formula to find the Base and height.
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Describe the transformation of the graph of f into the graph of g as either a horizontal or vertical stretch. f(x)=sqrt(x) and g
SpyIntel [72]
\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}\\
\left. \qquad   \right.  \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{  B}}\textit{ is negative}\\
\left. \qquad   \right.  \textit{reflection over the y-axis}

\bf \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}\\\\
\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}}}

with that template in mind, let's see    \bf f(x)=\sqrt{x}\qquad 
\begin{array}{llll}
g(x)=&\sqrt{0.5x}\\
&\quad \uparrow \\
&\quad  B
\end{array}

so B went form 1 on f(x), down to 0.5 or 1/2 on g(x)
B = 1/2, thus the graph is stretched by twice as much.
8 0
3 years ago
Read 2 more answers
(12x - 30) = (2x - 5)​
zzz [600]

Answer:

x = 5/2

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Algebra I</u>

  • Equality Properties

Step-by-step explanation:

<u>Step 1: Define</u>

12x - 30 = 2x - 5

<u>Step 2: Solve for </u><em><u>x</u></em>

  1. Subtract 2x on both sides:                    10x - 30 = -5
  2. Add 30 to both sides:                            10x = 25
  3. Divide 10 on both sides:                        x = 5/2
8 0
2 years ago
Is b = 6 a solution to the inequality b – 3 &gt; 5?
const2013 [10]

Answer:Hlello, my name is woods245.

So you have the problem b - 3>5 and is says that b=6.

So know the problem looks like this. 6 - 3>5.

Step-by-step explanation:

6 0
2 years ago
The average person's
OLga [1]

Answer:

20

Step-by-step explanation:

The answer would be 20 due to the fact that it would have to be 50 divided by 2.5 to get the answer of how many steps and 50 divided by 2.5 would equal 20

4 0
3 years ago
Two decimals that are equivalent to 3.200
AleksAgata [21]

Answer:

Two or more decimal numbers are said to be equivalent decimals, when they name the same value (or same amount). For example, the decimals: 0.2 = 0.20 = 0.200 = 0.2000 etc. Thus by successive addition of zeros after the decimal part of the number after the decimal point means the same number and hence are equivalent. Therefore, the decimals: 5.3, 5.30, 5.300, 5.3000 etc are all equivalent.

Step-by-step explanation:

Lol

5 0
2 years ago
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