Ask question

Ask question

All categories

Makovka662 [10]

4 years ago

15

Wich set of shapes contains members that are always similar to one another

2 answers:

Nadusha1986 [10]4 years ago

4 0

Hello.

The answer is

equilateral triangles
<span>All equilateral triangles are exactly the same as one another. They have the same size sides and the same shapes.
</span>
Have a nice day

iVinArrow [24]4 years ago

4 0

Answer: C) Equilateral Triangles

Any two triangles are similar if they have the same corresponding angles. In the case of two equilateral triangles, they both have 3 sets of 60 degree angles. So the angles pair up and we can prove them to be similar using the Angle Angle (AA) Similarity theorem.

This trick only works with triangles. It partially works with rectangles; however, we would need to be sure that the corresponding sides were proportional as well. There isn't a 100% guarantee that any two rectangles are similar.

You might be interested in

If the circumference of the circle is 37.68 units, what is the area? (Use 3.14 for pi .)

cricket20 [7]

Answer:

Area of the circle = $113.04\;units^2$

Step-by-step explanation:

Circumference = $37.68\;units$

Circumference of the circle = $2\times \pi \times r$

As,

$\pi =\dfrac{22}{7}=3.14$

$37.68=2\times \pi \times r\\\\37.68=2\times 3.14 \times r\\\\37.68=6.28\times r\\\\r=\dfrac{37.68}{6.28} \\\\r=6\;units$

Area of a circle = $\pi \times r^2$

$=3.14\times (6\times 6)\\\\=3.14\times 36\\\\=113.04\;units^2$

3 0

3 years ago

Solve x2 - 8x - 9 = 0.<br><br> Rewrite the equation so that it is of the form <br> x2 + bx = c.

stira [4]

X^2 - 8x-9=0
D = b^2- 4*a*c = 64- 4* (-9) = 64+36=100 =10^2
X1 = (8-10)/2 = -1
X2= (8+10)/2= 9

7 0

4 years ago

Find the error & find the correct answer 2In(x)=In(3x)-[In(9)-2In(3)] In(x^2)=In(3x)-[In(9)-In(9)] In(x^2)=In(3x)-0 In(x^2)=

Art [367]

Answer:

Error: $lnx^2=ln 3x$ not $ln\frac{3x}{0}$

Solution:x=0 and 3

Step-by-step explanation:

We have to find the error and correct answer

Given: $2ln x=ln(3x)-[ln9-2ln(3)]$

$lnx^2=ln(3x)-[ln9-ln3^2]$

Using the formula

$alog b=logb^a$

$lnx^2=ln(3x)-[ln9-ln9]$

$lnx^2=ln(3x)-0$

$lnx^2=ln(3x)$

$x^2=3x$

$x^2-3x=0$

$x(x-3)=0$

Therefore, x=0 and x=3

But last step in the given solution

$lnx^2=ln\frac{3x}{0}=\infty$

It is wrong this property is used when

$log m-log n$ then

$log\frac{m}{n}$

Hence, the student wrote $lnx^2=ln\frac{3x}{0}$ instead of $lnx^2=ln3x$ and solution is given by

x=0 and x=3

4 0

3 years ago

CAN SOMEONE CHECK MY WORK GEOMETRY PEOPLE ONLY PLS

quester [9]

Answer:

Is the graph shown the initial triangle, or the translated one?

Step-by-step explanation:

4 0

3 years ago

Help please inscribe angles

gladu [14]

Answer:

278°

Step-by-step explanation:

m arc BAC = 360 -(2×41)

= 360 - 82 = 278°

8 0

3 years ago