What is the surface area of the figure? A. 582 ft2 B. 291 ft2 C. 399 ft2 D. 900 ft2

Find the equation of a line that passes through the point (4,2) that is perpendicular to the line y=4/3 x

Mrac [35]

Answer:

$y=\frac{3}{4}x-1$

Step-by-step explanation:

Since we're looking at perpendicular lines, the slope of the new line is the reciprocal of the original line.

So the slope of our new line is $\frac{3}{4}$

So far, we are looking at $y=\frac{3}{4} x+b$

Now we need to find the b value by plugging in the given point.

$2=\frac{3}{4}(4)+b\\ 2=3+b\\ b=-1$

So the final equation is $y=\frac{3}{4}x-1$

8 0

3 years ago

between 1995 and 2006, the population of samburg, usa(in thousands) can be modeled by f(x) = 0.35x(squared) - 2.1x+15.8 where x=

Juli2301 [7.4K]

$\bf f(x)=0.35x^2-2.1x+15.8\qquad 
\begin{cases}
x=\textit{year since 1995}\\
f(x)=\textit{population amount}
\end{cases}$

the equation is a quadratic one, and it has a positive coefficient on the leading term, meaning, is opening upwards, so it has a "burrow" for the vertex.

the minimum or lowest point for a quadratic opening upwards is, well, the vertex point :), the "x" value is the year, the "y" or f(x) value is the population, we're asked for the year, or the x-coordinate of the vertex

well $\bf \begin{array}{llll}
f(x)=&0.35x^2&-2.1x&+15.8\\
&\quad \uparrow &\quad \uparrow&\uparrow \\
&\quad a&\quad b &c
\end{array}
\\\\

\\\\
\qquad \textit{vertex of a parabola}\\ \quad \\
\qquad 

\left(\boxed{-\cfrac{{{ b}}}{2{{ a}}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$