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4 years ago

Please help with geometry!!!!

Mathematics

1 answer:

AlladinOne [14]4 years ago

4 0

1) Parallelogram ABCD 1) Given

2) AD || BC; DC || AB 2) Definition of a parallelogram (opposite sides are ||)

3) <A + <B = 180 3) Same-side interior angles are supplementary

<B + <C = 180

4) <A and <B are supplementary 4) Definition of supplementary angles

<B and <C are supplementary

Hope this helps

You might be interested in

Square ABCD was translated using the rule (x, y) → (x – 4, y + 15) to form A'B'C'D'. What are the coordinates of point D in the

charle [14.2K]

X = 9+4 = 13
y = -8 - 15 = - 23

D(13,-23)

4 0

4 years ago

Read 2 more answers

What is the temperature in degrees Celsius that is equivalent to a temperature of 40 F?

Ludmilka [50]

Hi! So given the equation you are given, you plug in 40 for F, making your equation look like this: C=(5/9)(40-32). Using PEDMAS, Subtract 32 from 40 which gives you 8 and then multiply by 8. Your answer as an improper fraction should be 40/9 which you can then reduce to 4(4/9).

5 0

4 years ago

A student was asked to identify the following numbers as rational or irrational. Determine if the student has made any mistakes.

ioda

Answer:

Yes they are wrong, rational & irrational ones down below.

Step-by-step explanation:

So, lets start this by asking, what is a rational and irrational number?

Rational numbers are literally any numbers that arent pi, e, or other symbols, as well as square roots.

Irrational numbers are exactly the opposite, being the numbers that are symbols, including square roots.

Note that the symbols listed above such as pi are values that are neverending.

Now, using what we know of rational and irrational numbers, let see which answers are rational and which are not.

So, we start off with:

-13/3

I do not see a square root.

I do not see a symbol such as pi either.

It is not a ever going number.(number that continues infinitly)

<u>So -13/3 is a rational number.</u>

Next we have $\sqrt{37}$

immediatly we tell it has a square root.

<u>So </u> $\sqrt{37}$ <u> is a irrational number.</u>

Next we have 0.1234

This is not a every going number.

There isnt a square root.

And it isnt a symbol.

<u>So 0.1234 is a rational number.</u>

Next we have $-\sqrt{100}$

Again, we can immediatly know it has a square root.

<u>So </u> $-\sqrt{100}$ <u> is a irrational number.</u>

Next we have -77

This is not every going.

It doesnt have a square root.

It doesnt have a symbol.

So -77 is a rational number.

Lastly we have $-\sqrt{12}$

Like the other ones, we can imediatly tell it has a square root.

<u>So </u> $-\sqrt{12}$ <u> is a irrational number.</u>

Bascially, to answer further questions, here is the criteria for a number being rational:

It <u>does not</u> have a square root.
It <u>does not </u>have a symbol.
It<u> is not </u>a infinitly going on number.

If one of these criterias are not met, then you know its a irrational number.

Hope this helps! :D

6 0

3 years ago

Probe that:<br><img src="https://tex.z-dn.net/?f=%20%5Csec%20%5Calpha%20%20%5Csqrt%7B1%20-%20%20%5Csin%28%20%7B%7D%5E%7B2%7D%20%

Nataly [62]

Step-by-step explanation:

<h3> $\sec \alpha \sqrt{1 - \sin ^{2} \alpha } = 1$ </h3>

Prove the LHS

Using trigonometric identities

That's

<h3> $\cos ^{2} \alpha = 1 - \sin^{2} \alpha$ </h3>

<u>Rewrite the expression</u>

We have

<h3> $\sec \alpha \sqrt{ \cos^{2} \alpha }$ </h3>

<h3> $\sqrt{ { \cos }^{2} \alpha } = \cos \alpha$ </h3>

So we have

<h3> $\sec \alpha \times \cos \alpha$ </h3>

Using trigonometric identities

<h3> $\sec \alpha = \frac{1}{ \cos \alpha }$ </h3>

<u>Rewrite the expression</u>

That's

<h3> $\frac{1}{\cos \alpha } \times \cos \alpha$ </h3>

Reduce the expression with cos a

We have the final answer as

As proven

Hope this helps you

7 0

3 years ago

If four points are collinear, they are also coplanar. Apex -brainly<br> A. True<br> B. False

Tasya [4]

Answer:

B. False

Step-by-step explanation:

If four points are collinear, on the same line, then they don't have to be coplanar. Some of the points could be out of the plane, but still be collinear.

5 0

4 years ago