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3 years ago

PLZ HELP ME!!!! How can you find the areas of parallelograms, rhombus season, and trapezoids

Mathematics

1 answer:

Nikitich [7]3 years ago

3 0

Answer:

https://www.khanacademy.org/math/geometry/hs-geo-foundations/hs-geo-area/v/perimeter-and-area-basics

Step-by-step explanation:

Use the side bar

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CAN SOMEONE PLEASE HELP ME WITH THIS QUESTION?

Dimas [21]

Answer:

B. More than one quadrilateral exists with the given conditions, and all instances must be isosceles trapezoids.

Step-by-step explanation:

In a parallelogram, adjacent angles are supplementary. They are only congruent if the parallelogram is a rectangle. In this problem, adjacent angles are both congruent and acute. If this were a triangle, it would guarantee the triangle is isosceles.

The fact that opposite angles are supplementary guarantees that the fourth side of the figure is parallel to the base between the acute angles. That makes the figure an isosceles trapezoid. Unless specific angles and side lengths are specified, the description matches <em>any</em> isosceles trapezoid.

4 0

3 years ago

0/1 For positive integer n, n? = n! · (n − 1)! · … · 1! And n# = n? · (n − 1)? · … · 1?. What is the value of 4# · 3# · 2# · 1#?

MakcuM [25]

Answer:

331776

Step-by-step explanation:

Since n? = n! · (n − 1)! · … · 1! And n# = n? · (n − 1)? · … · 1?

Then 4# = 4? · (4 − 1)? · (4 − 2)?· 1?

= 4? · 3? · 2?· 1?

Now, n? = n! · (n − 1)! · … · 1!

So, 4? = 4! · (4 − 1)! · (4 − 2)! · 1! = 4! · 3! · 2! · 1! = 288

Thus, 3? = 3! · (3 − 1)! · 1! = 3! · 2! · 1! = 12

Also, 2? = 2! · (2 − 1)! · 1! = 2! · 1! · 1! = 2

and 1? = 1! · (1 − 1)! · 1! = 1! · 0! · 1! = 1

So, 4# = 4? · 3? · 2?· 1? = 288 × 12 × 2 × 1 = 6912

We now find 3#

3# = 3? · (3 − 1)? · 1? = 3? · 2?· 1?

Now, n? = n! · (n − 1)! · … · 1!

So, 3? = 3! · (3 − 1)! · 1! = 3! · 2! · 1! = 12

Thus, 2? = 2! · (2 − 1)! · 1! = 2! · 1! · 1! = 2

and, 1? = 1! · (1 − 1)! · 1! = 1! · 0! · 1! = 1

So, 3# = 3? · 2?· 1? = 12 × 2 × 1 = 24

We now find 2#

2# = 2? · (2 − 1)? · 1? = 2? · 1?· 1?

Now, n? = n! · (n − 1)! · … · 1!

So, 2? = 2! · (2 − 1)! · 1! = 2! · 1! · 1! = 2

1? = 1! · (1 − 1)! · 1! = 1! · 0! · 1! = 1

So, 2# = 2?· 1? = 2 × 1 = 2

We now find 1#

1# = 1? · 1? = 1? · 1?

Now, n? = n! · (n − 1)! · … · 1!

So, 1? = 1! · (1 − 1)! · 1! = 1! · 0! · 1! = 1

And, 1# = 1? · 1? = 1 × 1 = 1

So, 4# · 3# · 2# · 1#? = 6912 · 24 · 2 · 1? = 331776

7 0

2 years ago

Write an equation to represent the problem. Then solve the equation. Two years of local Internet service costs $685, including t

iren2701 [21]

Answer:

(Total - Fee) / Months = $Price per Month

(685 - 85) / 24 = 25

Step-by-step explanation:

4 0

3 years ago

Use substitution to solve the system of equations.

dusya [7]

Answer:

Step-by-step explanation:

1/2(2/3y) - 1/3y = 5

1/3y - 1/3y = 5

0 ≠ 5

No Solutions

These are linear equations and, when graphed, are parallel

7 0

3 years ago

Someone please help me out i don´t understand this question <br> ill give you 21 points

suter [353]

Answer:

0.67

Step-by-step explanation:

Formula: k = y/x

k = 2/3

k = 0.66666666666

k = 4/6

k= 0.66666666666

k = 6/9

k = 0.66666666666

k = 8/12

Round 0.67

Hence, answer = 0.67

[RevyBreeze]

8 0

2 years ago

Read 2 more answers