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3 years ago

Name the shapes that are being described.

1 answer:

6 0

A.) An equilateral triangle has three lines of symmetry. It has rotational symmetry of order 3. It has three equal sides.

b.) A Square (4 sides) has 4 Lines of Symmetry.

c.) A Regular Hexagon (6 sides) has 6 Lines of Symmetry

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Can you help me with #1?

steposvetlana [31]

The answer would be 27. If ABC is 3 times the length of AB then it would be the same for AC.

5 0

3 years ago

How many hours in 1 year?

oee [108]

8765.82 is s how many hours are in one year exactly

8 0

3 years ago

Read 2 more answers

image Alan, Benton, Claire, and Dita divided the cost of throwing a party as shown in the circle graph above. If Dita spent $ 12

sdas [7]

Let total amount spent =x

Dita spent 35 % of total amount.

35% of x = 0.35x

So amount spent by Dita = 0.35x

% of money spent by Benton = 20 % of x = 0.2x

We are given that Benton spent $12 more than Claire,

so making equation we have ,

Money spent by Dita = Money spent by Claire +12

$0.35x =0.2x +12$

$0.35x -0.2x = 12$

$0.5x =12$

x=80

So total amount spent was $80.

Money spent by Benton = 0.2x = 0.2 * 80= $16

So amount spent by Benton is $16.

5 0

4 years ago

What is the equation of the line that passes through the point (-4, 2) and has a<br> slope of ?

Katena32 [7]

Answer:

Step-by-step explanation:

What is the slope please

5 0

4 years ago

How do you integrate arctan(x dx? i think that if you simplify the integral you get:?

Temka [501]

Integration by parts will help here. Letting $u=\arctan x$ and $\mathrm dv=\mathrm dx$ , you end up with $\mathrm du=\dfrac{\mathrm dx}{1+x^2}$ and $v=x$ . Now

$\displaystyle\int\arctan x\,\mathrm dx=uv-\int v\,\mathrm du$
$\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\int\frac x{1+x^2}\,\mathrm dx$

For the remaining integral, setting $y=1+x^2$ gives $\dfrac{\mathrm dy}2=x\,\mathrm dx$ , so

$\displaystyle\int\frac x{1+x^2}\,\mathrm dx=\frac12\int\frac{\mathrm dy}y=\frac12\ln|y|+C=\frac12\ln(1+x^2)+C$

Putting everything together, you end up with

$\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\frac12\ln(1+x^2)+C$

6 0

3 years ago