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Keith_Richards [23]
3 years ago
10

Solve for x

7D%20" id="TexFormula1" title="0 = - { e }^{ - x} + 3 {e}^{3x} " alt="0 = - { e }^{ - x} + 3 {e}^{3x} " align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Brilliant_brown [7]3 years ago
8 0

\bf \textit{Logarithm Cancellation Rules} \\\\ \stackrel{\stackrel{\textit{we'll use this one}}{\downarrow }}{log_a a^x = x}\qquad \qquad a^{log_a x}=x~\hfill\stackrel{recall}{ln=log_e}\qquad log_e(e^z)=z \\\\[-0.35em] \rule{34em}{0.25pt}


\bf 0=-e^{-x}+3e^{3x}\implies e^{-x}=3e^{3x}\implies \cfrac{1}{e^x}=3e^{3x}\implies 1=e^x\cdot 3e^{3x} \\\\\\ 1=3e^xe^{3x}\implies \cfrac{1}{3}=e^{x+3x}\implies \cfrac{1}{3}=e^{4x}\implies ln\left( \cfrac{1}{3} \right)=ln\left( e^{4x} \right) \\\\\\ ln\left( \cfrac{1}{3} \right)=4x\implies \cfrac{ln\left( \frac{1}{3} \right)}{4}=x


and you plug that in your calculator to get about -0.27465307216702742285.

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Montano1993 [528]

Answer:

The answer to your question is:  L = 30 ft

Step-by-step explanation:

Data

man = 6 ft

man shadow = 5 ft

lamppost shadow = 25 ft

lamppost height = ? = L

Process

                   \frac{6}{5} = \frac{L}{25}

                    L = \frac{6 x 25}{5}

                    L = \frac{150}{5}

                           L = 30 ft

5 0
3 years ago
Ian measured a city park and made a scale drawing. In real life, the soccer field is 114 meters long. It is 19 millimeters long
ioda

Answer:

1 mm : 6 m

Step-by-step explanation:

Take the 19 mm and divide it by 114 meters. to get the scale

19/114  =  1/6

1 mm : 6 meters

8 0
3 years ago
Lesson: 1.08Given this function: f(x) = 4 cos(TTX) + 1Find the following and be sure to show work for period, maximum, and minim
Ber [7]

The given function is

f(x)=4\cos \text{(}\pi x)+1

The general form of the cosine function is

y=a\cos (bx+c)+d

a is the amplitude

2pi/b is the period

c is the phase shift

d is the vertical shift

By comparing the two functions

a = 4

b = pi

c = 0

d = 1

Then its period is

\begin{gathered} \text{Period}=\frac{2\pi}{\pi} \\ \text{Period}=2 \end{gathered}

The equation of the midline is

y_{ml}=\frac{y_{\max }+y_{\min }}{2}

Since the maximum is at the greatest value of cos, which is 1, then

\begin{gathered} y_{\max }=4(1)+1 \\ y_{\max }=5 \end{gathered}

Since the minimum is at the smallest value of cos, which is -1, then

\begin{gathered} y_{\min }=4(-1)+1 \\ y_{\min }=-4+1 \\ y_{\min }=-3 \end{gathered}

Then substitute them in the equation of the midline

\begin{gathered} y_{ml}=\frac{5+(-3)}{2} \\ y_{ml}=\frac{2}{2} \\ y_{ml}=1 \end{gathered}

The answers are:

Period = 2

Equation of the midline is y = 1

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Minimum = -3

3 0
1 year ago
Find the area of the figure. (Sides meet at right angles.)
daser333 [38]

Answer:

48m squared

Step-by-step explanation:

- cut the problem in half horizontally

you get two squares one square is 4 by 4 and the other 8 by 4

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6 0
3 years ago
Can you conclude from the figure that ABC is similar to DEF? Explain.​
pishuonlain [190]

Answer:

Yes, we can conclude that Triangle ABC is similar to triangle DEF because the measures of the 3 angles of both triangles are congruent.

Step-by-step explanation:

We have the measure of 2 angles from both triangles, and we know that triangles have 180°, so we can solve for the measure of the third angle for both triangles.

Triangle ABC:

Measure of angle A= 60°

Measure of angle C= 40°

Measure of angle B = 180°- (measure of angle A + measure of angle C) = 180° - (60° + 40°) = 80°

Triangle DEF

Measure of angle E= 80°

Measure of angle F= 40°

Measure of angle D= 180° - (measure of angle E + measure of angle F) = 180° - (80° + 40°) = 60°

The measures of the angles in Triangle ABC are: 60°, 40°, and 80°.

The measures of the angles in Triangle DEF are: 60°, 40°, and 80°.

Since the measure of 3 angles of the two triangles are the same, we know that the two triangles are similar.

8 0
3 years ago
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