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2 years ago

Which statements are true regarding the regular dodecagon? Check all that apply.

2 answers:

6 0

A dodecagon has 12 sides. That means that the smallest angle of rotational symmetry is 360/12 or 30°. All of its angles of rotational symmetry will be multiples of 30, including 180°. 135°, since it is not a multiple of 30°, is not a rotational angle for this figure. The order of °rotational symmetry is 12, as there are 12 sides you can turn this figure to (since it is regular).

Mumz [18]2 years ago

5 0

Which statements are true regarding the regular dodecagon? Check all that apply.

1) The smallest angle of rotational symmetry for the dodecagon is 30°.

2) The dodecagon has a rotational symmetry of 180°.

5) The angles of rotational symmetry for the dodecagon are multiples of 30°.

You might be interested in

Which of the following shows the extraneous solution to the logarithmic equation below???

avanturin [10]

Short Answer x = - 6
Remark
You gave the second best answer.

The trick here is not to divide both sides by 2. Solve the problem this way.

log_5(x + 1)^2 = 2 Take the antilog of both sides
(x + 1)^2 = 5^2 Expand the equation
x^2 + 2x + 1 = 25 Subtract 25 from both sides.
x^2 + 2x - 24 = 0 Factor
(x + 6)(x - 4) = 0 Find the zeros.

x + 6 = 0
x = - 6 <<<<<<<< Answer. This is the extraneous root.
The reason this is an extraneous root is that x<=0 do not have a logarithem

x - 4 = 0
x = 4 This is a legitimate result to the original equation.

7 0

2 years ago

Read 2 more answers

A survey found that 50% of teenagers prefer to watch a movie at a theater over other viewing options. You want to know the estim

ella [17]

Answer:

1. Flip a fair coin four times, with heads representing teenagers who prefer watching a movie at a theater and tails representing those who do not.

2. Pick a card from a deck of standard playing cards four times, with red suits representing those who prefer watching a movie at a theater and black suits representing those who do not.

4. Roll a six-sided die four times, with even numbers representing those who prefer watching a movie at a theater and odd numbers representing those who do not.

Step-by-step explanation:

Basically you need to do something 4 times, with each time having a 50/50 chance.

For 1, it flips the coin 4 times and a coin toss is 50/50

For 2, it pics 4 cards, and the deck of cards has a 50-50 chance of either red or black

For 3, it spins it only 3 times, so is wrong

For 4, it rolls the die 4 times, and even and odd is 50-50

For 5, 0-5 consist of 6 number, while 6-9 is only 4 numbers thus is not 50-50 so is wrong

6 0

2 years ago

Can somone give me the answer to 2 x 0

marshall27 [118]

Any number times 0 is going to equal 0

8 0

2 years ago

Read 2 more answers

The center of a circle is at the origin on a coordinate grid. The vertex of a parabola that opens upward is at (0, 9). If the ci

zhannawk [14.2K]

Answer:

"The maximum number of solutions is one."

Step-by-step explanation:

Hopefully the drawing helps visualize the problem.

The circle has a radius of 9 because the vertex is 9 units above the center of the circle.

The circle the parabola intersect only once and cannot intercept more than once.

The solution is "The maximum number of solutions is one."

Let's see if we can find an algebraic way:

The equation for the circle given as we know from the problem without further analysis is so far $x^2+y^2=r^2$ .

The equation for the parabola without further analysis is $y=ax^2+9$ .

We are going to plug $ax^2+9$ into $x^2+y^2=r^2$ for $y$ .

$x^2+y^2=r^2$

$x^2+(ax^2+9)^2=r^2$

To expand $(ax^2+9)^2$ , I'm going to use the following formula:

$(u+v)^2=u^2+2uv+v^2$ .

$(ax^2+9)^2=a^2x^4+18ax^2+81$ .

$x^2+y^2=r^2$

$x^2+(ax^2+9)^2=r^2$

$x^2+a^2x^4+18ax^2+81=r^2$

So this is a quadratic in terms of $x^2$

Let's put everything to one side.

Subtract $r^2$ on both sides.

$x^2+a^2x^4+18ax^2+81-r^2=0$

Reorder in standard form in terms of x:

$a^2x^4+(18a+1)x^2+(81-r^2)=0$

The discriminant of the left hand side will tell us how many solutions we will have to the equation in terms of $x^2$ .

The discriminant is $B^2-4AC$ .

If you compare our equation to $Au^2+Bu+C$ , you should determine $A=a^2$

$B=(18a+1)$

$C=(81-r^2)$

The discriminant is

$B^2-4AC$

$(18a+1)^2-4(a^2)(81-r^2)$

Multiply the (18a+1)^2 out using the formula I mentioned earlier which was:

$(u+v)^2=u^2+2uv+v^2$

$(324a^2+36a+1)-4a^2(81-r^2)$

Distribute the 4a^2 to the terms in the ( ) next to it:

$324a^2+36a+1-324a^2+4a^2r^2$

$36a+1+4a^2r^2$

We know that $a>0$ because the parabola is open up.

We know that $r>0$ because in order it to be a circle a radius has to exist.

So our discriminat is positive which means we have two solutions for $x^2$ .

But how many do we have for just $x$ .

We have to go further to see.

So the quadratic formula is:

$\frac{-B \pm \sqrt{B^2-4AC}}{2A}$

We already have $B^2-4AC}$

$\frac{-(18a+1) \pm \sqrt{36a+1+4a^2r^2}}{2a^2}$

This is t he solution for $x^2$ .

To find $x$ we must square root both sides.

$x=\pm \sqrt{\frac{-(18a+1) \pm \sqrt{36a+1+4a^2r^2}}{2a^2}}$

So there is only that one real solution (it actually includes 2 because of the plus or minus outside) here for x since the other one is square root of a negative number.

That is,

$x=\pm \sqrt{\frac{-(18a+1) \pm \sqrt{36a+1+4a^2r^2}}{2a^2}}$

means you have:

$x=\pm \sqrt{\frac{-(18a+1)+\sqrt{36a+1+4a^2r^2}}{2a^2}}$

$x=\pm \sqrt{\frac{-(18a+1)-\sqrt{36a+1+4a^2r^2}}{2a^2}}$ .

The second one is definitely includes a negative result in the square root.

18a+1 is positive since a is positive so -(18a+1) is negative

2a^2 is positive (a is not 0).

So you have (negative number-positive number)/positive which is a negative since the top is negative and you are dividing by a positive.

We have confirmed are max of one solution algebraically. (It is definitely not 3 solutions.)

If r=9, then there is one solution.

If r>9, then there is two solutions as this shows:

$x=\pm \sqrt{\frac{-(18a+1)+\sqrt{36a+1+4a^2r^2}}{2a^2}}$

r=9 since our circle intersects the parabola at (0,9).

Also if (0,9) is intersection, then

$0^2+9^2=r^2$ which implies r=9.

Plugging in 9 for r we get:

$x=\pm \sqrt{\frac{-(18a+1)+\sqrt{36a+1+4a^2(9)^2}}{2a^2}}$

$x=\pm \sqrt{\frac{-(18a+1)+\sqrt{36a+1+324a^2}}{2a^2}}$

$x=\pm \sqrt{\frac{-(18a+1)+\sqrt{(18a+1)^2}}{2a^2}}$

$x=\pm \sqrt{\frac{-(18a+1)+18a+1}{2a^2}}$

$x=\pm \sqrt{\frac{0}{2a^2}}$

$x=\pm 0$

$x=0$

The equations intersect at x=0. Plugging into $y=ax^2+9$ we do get $y=a(0)^2+9=9$ .

After this confirmation it would be interesting to see what happens with assume algebraically the solution should be (0,9).

This means we should have got x=0.

$0=\frac{-(18a+1)+\sqrt{36a+1+4a^2r^2}}{2a^2}$

A fraction is only 0 when it's top is 0.

$0=-(18a+1)+\sqrt{36a+1+4a^2r^2}$

Add 18a+1 on both sides:

$18a+1=\sqrt{36a+1+4a^2r^2$

Square both sides:

$324a^2+36a+1=36a+1+4a^2r^2$

Subtract 36a and 1 on both sides:

$324a^2=4a^2r^2$

Divide both sides by $4a^2$ :

$81=r^2$

Square root both sides:

$9=r$

The radius is 9 as we stated earlier.

Let's go through the radius choices.

If the radius of the circle with center (0,0) is less than 9 then the circle wouldn't intersect the parabola. So It definitely couldn't be the last two choices.

7 0

2 years ago

Read 2 more answers

Choose the correct answer. Two cars leave phoenix and travel along roads 90 degrees apart. If car 1 leaves 60 minutes earlier th

Alexxx [7]

Answer:

251.047804213 miles

Step-by-step explanation:

c1 t=3.5+1 speed 40 mph

c2 t=3.5 speed 50 mph

c1 40 *4.5= 180

c2 50 *3.5= 175

a^2+ b^2= c^2

180^2+175^2=c^2

32400+30625=c^2

63025=c^2

251.047804213=c

6 0

2 years ago