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d1i1m1o1n [39]
3 years ago
8

Is a kite a parallelogram?

Mathematics
1 answer:
Elena L [17]3 years ago
5 0

Answer: A kite is generally not considered a parallelogram. A kite is usually defined as having two sets of consecutive congruent sides. If the definition includes the phrase two DISTINCT sets of congruent sides it will not be a parallelogram, as the opposite sides will not be congruent.            

Step-by-step explanation:

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The cosine of 23° is equivalent to the sine of what angle
Archy [21]

Answer:

So 67 degrees is one value that we can take the sine of such that is equal to cos(23 degrees).

(There are more values since we can go around the circle from 67 degrees numerous times.)

Step-by-step explanation:

You can use a co-function identity.

The co-function of sine is cosine just like the co-function of cosine is sine.

Notice that cosine is co-(sine).

Anyways co-functions have this identity:

\cos(90^\circ-x)=\sin(x)

or

\sin(90^\circ-x)=\cos(x)

You can prove those drawing a right triangle.

I drew a triangle in my picture just so I can have something to reference proving both of the identities I just wrote:

The sum of the angles is 180.

So 90+x+(missing angle)=180.

Let's solve for the missing angle.

Subtract 90 on both sides:

x+(missing angle)=90

Subtract x on both sides:

(missing angle)=90-x.

So the missing angle has measurement (90-x).

So cos(90-x)=a/c

and sin(x)=a/c.

Since cos(90-x) and sin(x) have the same value of a/c, then one can conclude that cos(90-x)=sin(x).

We can do this also for cos(x) and sin(90-x).

cos(x)=b/c

sin(90-x)=b/c

This means sin(90-x)=cos(x).

So back to the problem:

cos(23)=sin(90-23)

cos(23)=sin(67)

So 67 degrees is one value that we can take the sine of such that is equal to cos(23 degrees).

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3 years ago
Regular hexagon FGHIJK shares a common shares a common center with square ABCD on a coordinate plane .
timurjin [86]

Answer:

<h3>"The answer is the yellow lines in the attached figure. </h3>

Step-by-step explanation: As shown in the attached figure, regular hexagon FGHIJK and square ABCD shares common centre on the co-ordinate plane and AB || FG.

We are to find the line across which the combined figure will reflect onto itself.

In the attached figure, we see two lines which are yellow in colour. We can easily detect that the figure will be reflected onto itself if these two lines acts as a mirror separately.

Hence these yellow lines are the required lines."

<h3><u>Hope this helps!</u></h3>

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Answer:

This app is meant for 1 question at a time

Step-by-step explanation:

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