Which graph represents the function y - 3 = 3/2(x – 4)?

1 answer:

5 0

There are no graphs attached, so I punched the equation into a calculator and this is what it gave me!! I hope this helps

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Nhich equation results from taking the square root of both sides of (x+9)2=25

ale4655 [162]

Answer:

square root of 25=5 and square root of (x+9)=(x+9) so the answer is.... x+9=+-5

Step-by-step explanation:

3 0

3 years ago

An older method of estimating body fat percentage was to measure a person's average density (total mass divided by total volume)

timofeeve [1]

If the density of the man is 1.0321 grams per cubic cm. Then the body fat percentage will be 28.59%.

<h3>What is a function?</h3>

The function is an expression, rule, or law that defines the relationship between one variable to another variable. Functions are ubiquitous in mathematics and are essential for formulating physical relationships.

An older method of estimating body fat percentage was to measure a person's average density (total mass divided by total volume) and apply a formula to convert it to a body fat percentage.

$\rm BF = \left (\dfrac{4.57}{p} - 4.142 \right) \times 100$

where BF = body fat and p = density in g/cm3 A man weighs 175 pounds.

A man weighs 175 pounds.

His total body volume is modeled by a cylinder with a height of 170 cm and a radius of 12 cm.

Then the density will be

$\rm p = \dfrac{175 \times 453.592}{\pi \times 12^2 \times 170}\\\\p = \dfrac{79378.7}{76906.19}\\\\p= 1.0321 \ g/cm^3$

Then the Body fat percentage will be

$\rm BF = \left (\dfrac{4.57}{1.0321} - 4.142 \right) \times 100 \\\\\\BF = 0.2859 \times 100\\\\\\BF = 28.59\%$

More about the function link is given below.

brainly.com/question/5245372

#SPJ1

8 0

2 years ago

What is the period and midline?

OverLord2011 [107]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\

\end{array}\qquad$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\
\bullet \textit{vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\


\end{array}$

$\bf \begin{array}{llll}
\bullet \textit{function period}\\
\qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\
\qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)
\end{array}
$

so if you notice yours $\bf \begin{array}{llll}
3.2cos&\left( \frac{5}{3}\theta \right)+&6.1\\
&\ \uparrow&\uparrow \\
&B&D 
\end{array}$

now.. normally the function $\bf 3.2cos&\left( \frac{5}{3}\theta \right)$
has a D value of 0, or no vertical shift, and the amplitude and the period simply make the wave taller and thinner, but the midline is still the x-axis

now, with D = 6.1, that moves the midline up vertically that much

now.. the period, well, B = 5/3, normal period of cosine is $2\pi$
so, the new period will be $\bf \cfrac{2\pi }{B}\implies \cfrac{2\pi }{\frac{5}{3}}\implies \cfrac{6\pi }{5}$

notice the picture below
the vertical shift by the D component, or 6.1, moved the midline to y = 6.1 :)