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3 years ago

HELP PLEASE Find the later area and total area

Mathematics

1 answer:

Mashcka [7]3 years ago

8 0

Answer:

LA=72

TSA= 72+24root3 (about 113.57)

Explanation:

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If f(x) is a linear function, f(−2)=4, and f(1)=−1, find an equation for f(x)

juin [17]

Answer:

$y = -\frac{5}{3}x+\frac{2}{3}$

Step-by-step explanation:

Given

$f(-2) = 4$

$f(1) = -1$

Required

The equation of the function

The given parameters means that:

$(x_1,y_1) = (-2,4)$

$(x_2,y_2) = (1,-1)$

Calculate the slope (m)

$m = \frac{y_2 -y_1}{x_2 -x_1}$

$m = \frac{-1-4}{1--2}$

$m = \frac{-5}{3}$

The equation is then calculated using:

$y = m(x - x_1) + y_1$

This gives:

$y =\frac{-5}{3}(x--2)+4$

$y =\frac{-5}{3}(x+2)+4$

Open bracket

$y = -\frac{5}{3}x-\frac{10}{3}+4$

Take LCM

$y = -\frac{5}{3}x+\frac{-10+12}{3}$

$y = -\frac{5}{3}x+\frac{2}{3}$

6 0

3 years ago

The graph of f(x) = x^2 has been shifted into the form f(x) = (x − h)^2 + k

stepan [7]

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

$\bf \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the y-axis}$

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

with that template in mind, let's see, it went to the right 2 units, and then up 3 units.

that simply means, C = -2, D = 3.

8 0

2 years ago

Why are we teching hligh levels of math at lower grades

Oksanka [162]

Answer:

I have no idea

Step-by-step explanation:

people are weird

8 0

3 years ago

Find the equation of the line

Valentin [98]