Ask question

Ask question

All categories

4 years ago

12

What is cos D?

2 answers:

Wittaler [7]4 years ago

8 0

It is A, Have a great day

Snowcat [4.5K]4 years ago

5 0

Cos D = adjacent of angle D/ Hypotenus
=33/65

You might be interested in

How do I rewrite the quadratic function in intercept

antiseptic1488 [7]

You have to factor and then get the x intercepts

8 0

3 years ago

What is the expression in simplest form? 21 - 6n + 2(2n -5 )-3n

Anit [1.1K]

31-7 because all you have to do is add the question

4 0

3 years ago

FAST REPLIES PLEASE I NEED HELP

Nonamiya [84]

Answer:

54 degrees

Step-by-step explanation:

No matter the categorization of a triangle the total degrees present will equal 180. You know that you have a corner that equals 44 degrees and another that is 82 degrees. Add the 44 + 82 degrees to get 126 degrees. After getting the 126 degrees, minus that from the total of 180 to get 54 degrees.

4 0

3 years ago

Solve the proportion 5/x=4/8 is

Anuta_ua [19.1K]

D. 10 is your answer.
In order to solve a proportion, you cross multiply.
In this case it will be 40=4x
Solve for x is 10. Therefore D is your answer.
Please give 5 star if it helped you. :)

7 0

3 years ago

Identify the polygon with vertices A(5,0), B(2,4), C(−2,1), and D(1,−3), and then find the perimeter and area of the polygon. HE

inn [45]

Answer:

Part 1) The polygon is a square

Part 2) The perimeter is equal to $20\ units$

Part 3) The area is equal to $25\ units^{2}$

Step-by-step explanation:

we have

$A(5,0), B(2,4), C(-2,1),D(1,-3)$

Plot the points

see the attached figure

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Find the distance AB

$A(5,0),B(2,4)$

substitute in the formula

$d=\sqrt{(4-0)^{2}+(2-5)^{2}}$

$d=\sqrt{(4)^{2}+(-3)^{2}}$

$d=\sqrt{25}$

$AB=5\ units$

Find the distance BC

$B(2,4), C(-2,1)$

substitute in the formula

$d=\sqrt{(1-4)^{2}+(-2-2)^{2}}$

$d=\sqrt{(-3)^{2}+(-4)^{2}}$

$d=\sqrt{25}$

$BC=5\ units$

Find the distance CD

$C(-2,1),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-1)^{2}+(1+2)^{2}}$

$d=\sqrt{(-4)^{2}+(3)^{2}}$

$d=\sqrt{25}$

$CD=5\ units$

Find the distance AD

$A(5,0),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-0)^{2}+(1-5)^{2}}$

$d=\sqrt{(-3)^{2}+(-4)^{2}}$

$d=\sqrt{25}$

$AD=5\ units$

we have that

AB=BC=CD=AD

Find the distance BD (diagonal)

$B(2,4),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-4)^{2}+(1-2)^{2}}$

$d=\sqrt{(-7)^{2}+(-1)^{2}}$

$BD=\sqrt{50}\ units$

<em>Verify if the polygon is a square</em>

If the triangle BDA is a right triangle, then the polygon is a square

Applying the Pythagoras theorem

$BD^{2}=AD^{2}+AB^{2}$

substitute

$(\sqrt{50})^{2}=5^{2}+5^{2}$

$50=50$ -----> is true

so

The triangle BDA is a right triangle

therefore

The polygon is a square

<em>Find the Area of the polygon</em>

The area of a square is equal to

$A=b^{2}$

we have

$b=5\ units$

$A=5^{2}=25\ units^{2}$

<em>Find the perimeter of the polygon</em>

The perimeter of a square is equal to

$P=4b$

we have

$b=5\ units$

$P=4(5)=20\ units$

6 0

3 years ago