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3 years ago

13

what are a shape with at least 1 pair of parallel sides, at least 1 pair of perpendicular sides, and only two right angles?

1 answer:

Jlenok [28]3 years ago

8 0

Answer:

trapeziods has 1 parallel sides

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7..4745 x 10 to the 10th power

Ludmilka [50]

Answer:

74,745,000,000

Step-by-step explanation:

First, you take the power and you move the decimal point over that man places:

7.4745 x 10 to the 10th power

Decimal point moves over 10 places to the right

= 74745,000,000

If the power was negative, you would move it left

Hope this helps!

Plz name brainliest if possible!

8 0

4 years ago

(will give brainliest!!)

Alex73 [517]

Hey there!

To figure out what "x" means, you can divide the numbers 126 and 14 . . .

126 divided by 14 = 9

9 x 14 = 126

9 = x

----------------
x = 9
----------------

Hope this helps you.
Have a great day!

4 0

3 years ago

Read 2 more answers

Match the expression with its name.

IrinaVladis [17]

Ax^2 + bx + c is a quadratic trinomial. Therefore, ur expression is a quadratic trinomial......it is a trinomial because it has 3 terms.....quadratic means the highest exponent is 2.

6 0

4 years ago

20 points! HELP AND ILL MARK BRAINLIEST!

Keith_Richards [23]

you do 77 divided by 14 you get 5.5 so the greatest number of teams is 5.

4 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