How do you solve 3x+9y=-7

2 answers:

8 0

Solve Equationssolve Sqrt[x - 1] + Sqrt[x + 1] == 2solve e^(2x) - 5e^x + 6 = 0 over the realssolve sin(x) + cos(2x) = 0Algebraevaluate x (2x-1) (2-x) at x = 3what is f(x) = (x^2 e^x)/(x-1) when x = -1Find Interceptsx intercepts of 2 x^3 - 5 x^2 + 4 x - 1y intercepts of 5x^4 -10x^2 + 2x - 2Medianmedian {-5,7,3,5,5,12}<span>
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Yuri [45]3 years ago

5 0

$3x+9y=-7\ \ \ \ \ |subtract\ 3x\ from\ both\ sides\\\\9y=-3x-7\ \ \ \ \ \ |divide\ both\ sides\ by\ 9\\\\\boxed{y=-\frac{1}{3}x-\frac{7}{9}}$

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What’s the answer and how do I solve it?

Dimas [21]

Answer:

x= -1, x=3

Step-by-step explanation:

The “zeros” of a quadratic function are where the parabola intersects the x axis. In that graph the parabola touches x=-1 and x=-3. Therefore that is the answer.

8 0

3 years ago

Find an equation for the line perpendicular to y=−15x+3 with x-intercept at x = 3.

Alex Ar [27]

bearing in mind that perpendicular lines have negative reciprocal slopes, let's find the slope of the provided line then

$\bf y=\stackrel{\stackrel{m}{\downarrow }}{-15}x+3\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill$

$\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-15\implies -\cfrac{15}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{1}{15}}\qquad \stackrel{negative~reciprocal}{+\cfrac{1}{15}\implies \cfrac{1}{15}}}$

well, we know the x-intercept is at x = 3, recall when a graph intercepts the x-axis y = 0, so this point is (3 , 0). Then we're really looking for the equation of a line whose slope is 1/5 and runs through (3 , 0).

$\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{0})~\hspace{10em} slope = m\implies \cfrac{1}{5} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-0=\cfrac{1}{5}(x-3)\implies y=\cfrac{1}{5}x-\cfrac{3}{5}$