Determine if the two figures are congruent and explain your answer.

Mathematics

1 answer:

Ne4ueva [31]3 years ago

7 0

They arent congruent when you look at the side lengths and area

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AActiveWhich is the graph of g(x)=[²]* - 2²(0,2.25)2--2Mark this and return2 3Save and Exit56NextSubmit

bogdanovich [222]

SOLUTION

Given the question in the image, the following are the solution steps to answer the question.

STEP 1: Write the given function

$g(x)=(\frac{2}{3})^x-2$

STEP 2: Plot the given function

STEP 3: choose the correct graph

It can be seen from the plotted graph in step 2 that:

$\begin{gathered} x-intercept:(-1.71,0) \\ y-intercept:(0,-1) \end{gathered}$

Hence, comparing the x-intercepts, the correct graph is seen in:

3 0

1 year ago

S^2t - 10 if s= -8 and t = 3/4

mart [117]

Answer:

-58

Step-by-step explanation:

input

-8^2(3/4)-10

simplify

-8^1.5-10

then solve

3 0

3 years ago

Read 2 more answers

3y''-6y'+6y=e*x sexcx

Simora [160]

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

$3r^2-6r+6=0\iff r^2-2r+2=0$

which has roots at $r=1\pm i$ . This admits the two fundamental solutions

$y_1=e^x\cos x$
$y_2=e^x\sin x$

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

$u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx$
$u_2=\dfrac13x$

Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$