Find the value of A, B, C and D

Mathematics

1 answer:

bekas [8.4K]3 years ago

5 0

Since some sides are congruent to the mirrored of itself, you can conclude that a = 65 degrees
b = 180 - 65 = 115 degrees
c = 20 degrees (180 = 45 + 115 + x)
d = 45 degrees

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An exponential growth function has an asymptote of y = –3. Which might have occurred in the original function to permit the rang

Blababa [14]

The right choice is: A whole number constant could have been subtracted from the exponential expression.

Let be an exponential function of the form $y = A\cdot e^{B\cdot x}$ , where $A$ and $B$ are real numbers. A horizontal asymptote exists when $e^{B\cdot x} \to 0$ , which occurs for $B\cdot x \to - \infty$ .

For this function, the horizontal asymptote is represented by $y = 0$ and to change the value of the asymptote we must add the parent function by another real number ( $C$ ), that is to say:

$y = A\cdot e^{B\cdot x} + C$ (1)

In this case, we must use $C = -3$ to obtain an horizontal asymptote of -3. Thus, the right choice is: A whole number constant could have been subtracted from the exponential expression.

To learn more on asymptotes, we kindly invite to check this verified question: brainly.com/question/8493280

3 0

2 years ago

Read 2 more answers

How do you do this problem? I need to know how you found the answer.

Alexxandr [17]

to get the equation of a line, we simply need two points, say for the Red one ... notice in the graph the lines passes through (0,2) and (-1,6), so let's use those

$\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{-1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-2}{-1-0}\implies \cfrac{4}{-1}\implies -4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=-4(x-0) \\\\\\ y-2=-4x\implies \blacktriangleright y=-4x+2 \blacktriangleleft$

now, for the Blue one, say let's use hmmm it passes through (0,2) and (1.6)

$\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-2}{1-0}\implies \cfrac{4}{1}\implies 4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=4(x-0) \\\\\\ y-2=4x\implies \blacktriangleright y=4x+2 \blacktriangleleft$