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2 years ago

Can a rectangle be a parallelogram

1 answer:

7 0

A parallelogram is a quadrilateral with two pairs of opposite, equal in length and is parallel to one another.

A rectangle is a type of parallelogram.

So, yes, a rectangle is a parallelogram.

You might be interested in

Write a number with the digit 8 in the tens place

Anon25 [30]

80 because tens place is two over from the right

8 0

3 years ago

Read 2 more answers

Find the range of F<img src="https://tex.z-dn.net/?f=f%28x%29%3D%20%5Cfrac%7B2%7D%7Bx-7%7D%20" id="TexFormula1" title="f(x)= \fr

steposvetlana [31]

Answer: $y \in R; y \neq 0$
Explanation: For this, it is often best to find the horizontal asymptote, and then take limits as x approaches the vertical asymptote and the end behaviours.

Well, we know there will be a horizontal asymptote at y = 0, because as x approaches infinite and negative infinite, the graph will shrink down closer and closer to 0, but never touch it. We call this a horizontal asymptote.
So we know that there is a restriction on the y-axis.

Now, since we know the end behaviours, let's find the asymptotic behaviours.

As x approaches the asymptote of 7⁻, then y would be diverging out to negative infinite.
As x approaches the asymptote at 7⁺, then y would be diverging out to negative infinite.

So, our range would be: $y \in R; y \neq 0$

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7Bx%20%5Csqrt%7Bx%20%5Csqrt%7Bx%20%5Csqrt%7Bx....%7D%20%7D%20%7D%20%7D%20%20%3D%20%

andrey2020 [161]

First observe that if $a+b>0$ ,

$(a + b)^2 = a^2 + 2ab + b^2 \\\\ \implies a + b = \sqrt{a^2 + 2ab + b^2} = \sqrt{a^2 + ab + b(a + b)} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{\cdots}}}}$

Let $a=0$ and $b=x$ . It follows that

$a+b = x = \sqrt{x \sqrt{x \sqrt{x \sqrt{\cdots}}}}$

Now let $b=1$ , so $a^2+a=4x$ . Solving for $a$ ,

$a^2 + a - 4x = 0 \implies a = \dfrac{-1 + \sqrt{1+16x}}2$

which means

$a+b = \dfrac{1 + \sqrt{1+16x}}2 = \sqrt{4x + \sqrt{4x + \sqrt{4x + \sqrt{\cdots}}}}$

Now solve for $x$ .

$x = \dfrac{1 + \sqrt{1 + 16x}}2 \\\\ 2x = 1 + \sqrt{1 + 16x} \\\\ 2x - 1 = \sqrt{1 + 16x} \\\\ (2x-1)^2 = \left(\sqrt{1 + 16x}\right)^2$

(note that we assume $2x-1\ge0$ )

$4x^2 - 4x + 1 = 1 + 16x \\\\ 4x^2 - 20x = 0 \\\\ 4x (x - 5) = 0 \\\\ 4x = 0 \text{ or } x - 5 = 0 \\\\ \implies x = 0 \text{ or } \boxed{x = 5}$

(we omit $x=0$ since $2\cdot0-1=-1\ge0$ is not true)

3 0

1 year ago

THIS IS 99 POINTS!!!!!!! WILL REPORT IF YOU DONT ANSWER CORRECTLY

Umnica [9.8K]

Gg easey

perimiter is jsut the sum of the measures of the sides

P=2.3x+14+2x-0.2x+15
group like terms
P=2.3x+2x-0.2x+14+15
add
P=4.1x+29

answer is first one

4 0

3 years ago

Read 2 more answers

Please solve the following question.

mylen [45]

Answer:

103, 776

Step-by-step explanation:

There are 48 possibilities, and you are picking 3.

Now, the order does matter (aka, if the same 3 athletes won but got different places, we would still consider it to be a separate possibility)

(this is considered "without repetitions")

the number of permutations (arrangement combinations where the order does matter) without repetitions formula:

Number of permutations

without <em>repetitions </em>= nPr

= $\frac{n!}{(n-r)!}$

(

[the P is for combinations [where the order does matter, if the order didn't matter then it would be C, and it would be a different formula.]

n is the total number of objects.

r is the number of objects selected]

<em>(a repetition would be like two versions of the same person competing, which doesn't make sense)</em>

{the ! means factorial:

5! = 5 x 4 x 3 x 2 x 1}

{for example, if there are 5 people to give a presentation, and they can go in any order [but cannot repeat their presentation], so they all must fill 5 slots}

{for the first slot there are 5 choices, the second slot there are 4 choices...}

)

So, if we follow this formula:

$\frac{n!}{(n-r)!}$

n: 48

r: 3

$\frac{48!}{(48-3)!}$

= $\frac{48!}{45!}$

= $\frac{48 * 47*46*45!}{45!}$

(the 45! cancel out)

= 48 * 47 * 46

= 103, 776

(or, without the formula:

for the first choice (lets say of gold), there are 48 options

for the second choice (lets say silver), there are 47 options left to choose from

for the third choice (lets say bronze), there are 46 options left to choose from

)

3 0

2 years ago