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3 years ago

50 points! Congruent triangles!

Mathematics

2 answers:

aleksandr82 [10.1K]3 years ago

7 0

Answer: Rotated 90° about the origin and reflected across the x-axis

Why? Because dilation with a scale factor different from 1 makes the figure similar to another.

Similar figures are congruent only when the similarity scale factor is equal to 1.

sasho [114]3 years ago

3 0

“Rotated 90 about the origin and reflected across the x-axis”

Because to be congruent, there must not be any dilation

Hope this helps

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A standard container measures 2.4 m by 2.4 m by 6 m.

ella [17]

Answer:

34.56

Step-by-step explanation:

5 0

3 years ago

Beverley has quarters and dimes in her pocket. She has 21 coins with a total value of $3.00. How many of each type of coin does

kondaur [170]

Answer:

Step-by-step explanation:

If all 21 coins were dimes, she'd have $2.10. But she has 90 more cents than that. Now for each coin that's a quarter, not a dime, she has 15 cents more. 90/15= 6. So there are 6 quarters, and 21-6 = 15 dimes.

Now we do it backwards to see if it's right. 6 quarters is $1.50, 15 dimes is $1.50, add those together and you get $3. So it's right. 6 quarters, 15 dimes.

6 0

3 years ago

You should be able to find this

Nastasia [14]

Answer:

$\frac{1}{2}$

Step-by-step explanation:

sin( $cos^{-1}$ $\frac{\sqrt{3} }{2}$ )

using the sides of the 30- 60- 90 triangle

with legs 1, $\sqrt{3}$ and hypotenuse 2 , then

$cos^{-1}$ $\frac{\sqrt{3} }{2}$ = 30° and

sin30° = $\frac{1}{2}$

3 0

2 years ago

Which statement describes if there is an extraneous solution to the equation √x-3 = x-5? A. there are no solutions to the equati

Brrunno [24]

Remember that an extraneous solution of an equation, is the solution that emerges from solving the equation but is not a valid solution.

Lets solve our equation to find out what is the extraneous solution:
 $\sqrt{x-3} =x-5$
$(\sqrt{x-3})^2 =(x-5)^2$
$x-3=x^2-10x+25$
$x^2-11x+28=0$
$(x-4)(x-7)=0$
$x-4=0$ and $x-7=0$
$x=4$ and $x=7$

So, the solutions of our equation are $x=4$ and $x=7$ . Lets replace each solution in our original equation to check if they are valid solutions:
- For $x=7$
$\sqrt{x-3} =x-5$
$\sqrt{7-3} =7-5$
$\sqrt{4} =2$
$2=2$
We can conclude that 7 is a valid solution of the equation.

- For $x=4$
$\sqrt{x-3} =x-5$
$\sqrt{4-3} =4-5$
$\sqrt{1} =1$
$1 \neq 1$
We can conclude that 4 is not a valid solution of the equation; therefore, 4 is a extraneous solution.

We can conclude that the correct answer is: D. the extraneous solution is x = 4

7 0

3 years ago

Help me please I will give brainliest

In-s [12.5K]

Answer:

What grade is this? I may have an answer.

Step-by-step explanation:

7 0

3 years ago