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3 years ago

Quadrilateral ABCD has coordinates A(3,5) B(5,2) C(8,4) D(6,7). quadrilateral ABCD is a?

2 answers:

7 0

Use the distance formula. The distance formula is:

$\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

If you input the values, you will get:

A to B= $\sqrt{13}$

B to C= $\sqrt{13}$

C to D= $\sqrt{13}$

D to A= $\sqrt{13}$

It is a square.

Hope that helped!

~Cam943, Moderator

Lorico [155]3 years ago

4 0

Answer:

it is a square use geogebra and you will see

Step-by-step explanation:

You might be interested in

For which function is the ordered pair (2, 12) not a solution?

Ksju [112]

Answer:

option b

Step-by-step explanation:

y=12 x=2

12 is not equal to 2-10.

lemme know if you didn't understand

7 0

3 years ago

Use The Alternate Interior Angles Theorem diagram to answer the question. Give the missing reason in this proof for the letter g

Serjik [45]

The missing reasons in the proof using the Alternate Interior Angles Theorem diagram that is given are:

a. Corresponding angles

b. Vertical Angles, and

c. Alternate interior angles

Two angles lying opposite each other along a transversal, and are within the two lines crossed by the transversal are referred to as alternate interior angles.

According to the Alternate Interior Angles Theorem, the two alternate interior angles are congruent to each other.

Let's state out our proof using the image given:

Statement 1: line l is parallel to line m

Reason: Given

Statement 2: $\angle 2 \cong \angle 6$

Reason: Corresponding Angles (both angles occupy the same corner, hence they correspond to each other. Corresponding angles have the same measure).

Statement 3: $\angle 4 \cong \angle 2$

Reason: Vertical angles (both angles are directly opposite each other as they share the same point, which makes them vertical angles. Vertical angles have equal measure).

Statement 4: $\angle 6\cong \angle 4$

Reason: Alternate Interior Angles (applying the transitive property which says if a = b, and b = c, then a = c, therefore, since $\angle 2 \cong \angle 6, $ and $ \angle 4 \cong \angle 2, $ then $ \angle 6 \cong \angle 4$ )

In conclusion, the missing reasons in the proof using the Alternate Interior Angles Theorem diagram that is given are:

a. Corresponding angles

b. Vertical Angles, and

c. Alternate interior angles

Learn more here:

brainly.com/question/18898800

4 0

2 years ago

Three runners run along a 1.5 mile trail one Saturday morning. The graph shows the runners’ locations starting at 9:00 a.m.

Nostrana [21]

Answers:

The graph is attached below. With this information we have to answer the questions:

1) Who is at the trail start at 9:00 a.m.?

Andy

According to the graph, at 9:00 a.m. Andy is at the point (0,0), which is the begining of the trail. Hence, he is at the trail start.

2) Who runs fastest?

Bertha

If we observe the slopes of each line; Bertha's slope has the least steep. This means she is displacing more in less time, than the other two runners.

3) When do Andy and Berta cross paths?

After 8 minutes, on mile 1.

According to the graph, exactly at the point (8,1), which is at 9:08 a.m. and in mile 1; is when the paths of both runners intercept each other.

4) Who is closest to the 1.5 mile mark after 6 minutes?

Bertha

As we see in the graph, at 9:06 a.m Bertha is closest to the 1.5 mile mark, since her line is closest to this point at that time.

4 0

3 years ago

Read 2 more answers

Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

6 0

3 years ago

Help! please!!!!!! look at photo :))

JulsSmile [24]

Hey there!

We know that Danielle earns $10 per hour, so muliply that by 3 and get 30.

Because Danielle works an extra half an hour, divide 10 by 2 and get 5.

Danielle earns $35 in 3 hours and a half.

Hope this helps! Please mark me as brainliest!

Have a wonderful day :)

4 0

3 years ago