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Anit [1.1K]
3 years ago
15

Use the graphing tool to see what happens to the logarithmic (log base 10) function when you change more than one parameter at a

time. In what ways is this new graph different from the parent graph of f(x)=log(x)? Write down your observations for each of these transformed functions. Then try a few of your own.
Mathematics
1 answer:
ivann1987 [24]3 years ago
5 0

Answer:

The parent graph of g(x) = 5 log (x + 1) shifts one unit to the left and is stretched vertically by a factor of 5 .

The parent graph of g(x) = 1/5 log (1/7 x)   is stretched horizontally by a factor of 7  and then compressed vertically by a factor of 1/5 .

Step-by-step explanation:

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The equation of line WX is 2x + y = -5. What is the equation of a line perpendicular to line WX in slope-intercept form that con
sveticcg [70]

Answer:

The third choice is the one you want

Step-by-step explanation:

If we are to write the equation of a line perpendicular to WX, we first must determine what the slope of the WX is, because the line perpendicular to WX has a slope that is the flip of the slope of WX with the opposite sign.  Solving for y takes care of finding the slope of WX:

2x + y = -5 so

y = -2x - 5

The slope is -2.  That means that the reciprocal slope is 1/2.  Using that slope along with the coordinates x = -1 and y = -2, we first write the line using point-slope form and then solve it for y.  Start by filling in the m, the x value and the y value:

y - (-2) = \frac{1}{2}(x-(-1))

Getting rid of the double negatives gives us:

y+2=\frac{1}{2}(x+1)

Distributing then gives us:

y+2=\frac{1}{2}x+\frac{1}{2}

And finally solving for y (I am going to express the 2 on the left as 4/2 when I move it by subtraction in order to add those fractions):

y=\frac{1}{2}x+\frac{1}{2}-\frac{4}{2}

And the final equation in slope-intercept form is:

y=\frac{1}{2}x-\frac{3}{2}

8 0
3 years ago
PLEASE HELP I CANT CONTINUE WITH MY WORK UNLESS THESE ARE COMPLETE AND CORRECT
Snezhnost [94]

Answer:

1) x=3, y=-3

Step-by-step explanation:

10x + 7y = 9 —— (1)

-4x -7y = 9 —— (2)

(1) + (2)

6x = 18

x = 18/6

x = 3

put x=3 in (1)

10(3) + 7y = 9

30 + 7y = 9

7y = 9 - 30

7y = -21

y = -21/7

y = -3

please give me brainlist if it is helpful for you

6 0
3 years ago
Write each equation in standard form. identify A,B,C.
aleksley [76]
Write in y=ax²+bx+c form
solve for y

given
\frac{x+5}{3}=-2y+4
solve for y

easy
minus 4 both sides
\frac{x+5}{3}-4=-2y
times -1/2 to both sides
\frac{x+5}{-6}+2=y
y=\frac{x+5}{-6}+2
y=\frac{-x}{6}-\frac{6}{5}+2
2=10/5
y=\frac{-1}{6}x-\frac{6}{5}+\frac{10}{5}
y=\frac{-1}{6}x+\frac{4}{5}
y=ax²+bx+c
y=0x^2+\frac{-1}{6}x+\frac{4}{5}

a=0
b=\frac{-1}{6}
c=\frac{4}{5}
3 0
3 years ago
-x+3=-3d-3r, for x .
OLga [1]

Answer:

x=3d+3r+3

Step-by-step

5 0
3 years ago
The points A(1, 4), B(5,1) lie on a circle. The line segment AB is a chord. Find the equation of a diameter of the circle.
tangare [24]

Check the picture below.

well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.

bearing in mind that <u>perpendicular lines have negative reciprocal</u> slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.

\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}

so, it passes through the midpoint of AB,

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{5+1}{2}~~,~~\cfrac{1+4}{2} \right)\implies \left(3~~,~~\cfrac{5}{2} \right)

so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)

\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{\frac{5}{2}}) \stackrel{slope}{m}\implies \cfrac{4}{3} \\\\\\ \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-\stackrel{y_1}{\cfrac{5}{2}}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{3})\implies y-\cfrac{5}{2}=\cfrac{4}{3}x-4 \\\\\\ y=\cfrac{4}{3}x-4+\cfrac{5}{2}\implies y=\cfrac{4}{3}x-\cfrac{3}{2}

4 0
3 years ago
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