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jeka57 [31]
2 years ago
14

Which pairs of triangles can be shown to be congruent using rigid motions?

Mathematics
2 answers:
In-s [12.5K]2 years ago
7 0

Here is the answers hope this helps enjoy  

Reika [66]2 years ago
4 0

Answer:

1. Not congruent

2. Not congruent

3. Congruent

4. Congruent

Step-by-step explanation:

Accroding to the options:

1. ΔABC is not congruent to ΔMPN.

As, we can see that the distance between AC and the point B is 5 units. If we rotate ΔABC around the point B downwards and then translate it downwards.

We may obtain ΔMPN but, in ΔMPN the distance between MN and point P is 4 units.

So, ΔABC will not be congruent to ΔMPN.

2. ΔABC is not congruent to ΔEGF.

As, we can see that the distance between AC and the point B is 5 units. If we rotate ΔABC around the point B upward and then translate it downwards.

We may obtain ΔEGF but, in ΔEGF the distance between EF and point G is 4 units.

So, ΔABC is not congruent to ΔEGF.

3. ΔABC is congruent to ΔSTU.

As, we can see that the distance between AC and the point B is 5 units. If we reflect ΔABC around the line AC and then translate it 5 units to left. We obtain ΔSTU.

So, ΔABC is congruent to ΔSTU.

3. ΔEFG is congruent to ΔMNP.

As, we can see that the distance between EF and the point G is 4 units. If we reflect ΔEFG around the line EF and then translate it 4 units to downwards. We obtain ΔMNP.

So, ΔEFG is congruent to ΔMNP.

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PLEASE!!! NEED HELP ASAP!!! 50 PTS AND BRAINLIEST!!!
noname [10]

1) Inverse function: f^{-1}(x)=g(x)=\frac{x+4}{3}

2) f(1) = -1, g(-1) = 1

3) f(g(x))=g(f(x))=x

Step-by-step explanation:

1)

In this first method, we want to find the inverse function of f(x).

The original function is:

f(x) = 3x-4

We rewrite it as

y=3x-4

We swap the name of the variables:

x=3y-4

Adding +4 on both sides:

x+4=3y-4+4\\x+4=3y

And dividing by 3 on both sides:

y=\frac{x+4}{3}

which is identical to g(x):

g(x)=\frac{x+4}{3}

2)

In this second method, we want to use the output of one function as input to the other function, and show that the output value is equal to the imput value.

We start using the function

f(x)=3x-4

We choose x=1. We find:

f(1)=3(1)-4=3-4=-1

Now we use this output value as input in the function

g(x)=\frac{x+4}{3}

Substituting x=-1,

g(-1)=\frac{-1+4}{3}=\frac{3}{3}=1

So, the final output value (1) is equal to the input value (1).

3)

Here we want to verify that the two are inverse functions by showing that

f(g(x))=x

We have

f(x)=3x-4\\g(x)=\frac{x+4}{3}

Substituting g(x) into f(x),

f(g(x))=3g(x)-4 = 3(\frac{x+4}{3})-4=(x+4)-4=x

Viceversa, we want to show that

g(f(x))=x

Substituting g(x) into f(x), we get

g(f(x))=\frac{f(x)+4}{3}=\frac{(3x-4)+4}{3}=\frac{3x-4+4}{3}=\frac{3x}{3}=x

So, the two functions are one the inverse of the other.

Learn more about inverse functions:

brainly.com/question/1632445

brainly.com/question/2456302

brainly.com/question/3225044

#LearnwithBrainly

6 0
3 years ago
Plz answer i will really apreciate it :) 50 points
hoa [83]

Answer:

b or c

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
The freshmen class is going on a field trip. There are 316 people going.
marishachu [46]
The equation would be 6x+5y=316. X represents the car travelers and the y represents the bus travelers
3 0
3 years ago
Each day at lunchtime, at least 53 people buy food from a food truck.
Nataly_w [17]

Answer:

p ≥ 53

Step-by-step explanation:

6 0
3 years ago
Find the average rate of change of the function over the given interval
sattari [20]
\bf slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ f(x_2)}}-{{ f(x_1)}}}{{{ x_2}}-{{ x_1}}}\impliedby 
\begin{array}{llll}
average\ rate\\
of\ change
\end{array}\\\\
-------------------------------\\\\

\bf h(t)=cot(t)\implies h(t)=\cfrac{cos(t)}{sin(t)}\quad 
\begin{cases}
t_1=\frac{\pi }{4}\\
t_2=\frac{3\pi }{4}
\end{cases}\implies \cfrac{h\left( \frac{3\pi }{4} \right)-h\left( \frac{\pi }{4} \right)}{\frac{3\pi }{4}-\frac{\pi }{4}}
\\\\\\

\bf \cfrac{\frac{cos\left( \frac{3\pi }{4} \right)}{sin\left( \frac{3\pi }{4} \right)}-\frac{cos\left( \frac{\pi }{4} \right)}{sin\left( \frac{\pi }{4} \right)}}{\frac{\pi }{2}}\implies \cfrac{-1-1}{\frac{\pi }{2}}\implies \cfrac{-2}{\frac{\pi }{2}}\implies -\cfrac{4}{\pi }\\\\\\
-------------------------------\\\\

\bf h(t)=cot(t)\implies h(t)=\cfrac{cos(t)}{sin(t)}\quad 
\begin{cases}
t_1=\frac{\pi }{3}\\
t_2=\frac{3\pi }{2}
\end{cases}\implies \cfrac{h\left( \frac{3\pi }{2} \right)-h\left( \frac{\pi }{3} \right)}{\frac{3\pi }{2}-\frac{\pi }{3}}
\\\\\\

\bf \cfrac{\frac{cos\left( \frac{3\pi }{2} \right)}{sin\left( \frac{3\pi }{2} \right)}-\frac{cos\left( \frac{\pi }{3} \right)}{sin\left( \frac{\pi }{3} \right)}}{\frac{9\pi -2\pi  }{6}}\implies \cfrac{\frac{0}{-1}-\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}}{\frac{7\pi }{6}}\implies\cfrac{-\frac{1}{\sqrt{3}}}{\frac{7\pi }{6}}\implies -\cfrac{\sqrt{3}}{3}\cdot \cfrac{6}{7\pi }
\\\\\\
-\cfrac{2\sqrt{3}}{7\pi }
8 0
3 years ago
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