Are these parallelograms or not?

Find the distance between the points (1,8) and (9, 2).

grin007 [14]

Answer:

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

$=\sqrt{(9-1)^2+(2-8)^2 }\\\\=\sqrt{8^2 + 6^2}\\\\=\sqrt{64 +36 }\\\\=\sqrt{100} = 10units$

6 0

3 years ago

Which pattern was created using only the translation of pentagon A?

Makovka662 [10]

Translation means only sliding it around, no rotation is allowed. So it’s the first one.

8 0

2 years ago

7 1/6 = 5 2/5 + p<br><br> p = ???

Crank

$\huge\textbf{Hey there!}$

$\mathbf{7\dfrac{1}{6} = 5\dfrac{2}{5} + p}$

$\rightarrow\mathbf{7 \dfrac{1}{6} = 5\dfrac{2}{5} + p}$

$\mathbf{\rightarrow p + 5\dfrac{2}{5}= 7\dfrac{1}{6}}$

$\mathbf{\rightarrow {p + \dfrac{27}{5}=\dfrac{43}{6}}}$

$\text{SUBTRACT }\rm{\dfrac{27}{5}}\text{ to BOTH SIDES}$

$\mathbf{p + \dfrac{27}{5} - \dfrac{27}{5} = \dfrac{43}{6}- \dfrac{27}{5}}$

$\text{CANCEL out: }\rm{\dfrac{27}{5} - \dfrac{27}{5}}\text{ because it gives you 0}$

$\text{KEEP: }\rm{\dfrac{43}{6} - \dfrac{27}{5}}\text{ because it help solve for the p-value}$

$\text{NEW EQUATION: }\rm{p = \dfrac{43}{6} - \dfrac{27}{5}}$

$\large\textsf{SIMPLIFY IT!}$

$\mathbf{p = \dfrac{53}{30} \approx p = 1\dfrac{23}{30}}$

$\huge\boxed{\textsf{Therefore, your answer is: \boxed{\mathsf{p = 1\dfrac{23}{30}}}}}\huge\checkmark$

$\huge\textbf{Good luck on your assignment \&}\\\huge\textbf{enjoy your day!}$

~ $\frak{Amphitrite1040:)}$

3 0

2 years ago

Read 2 more answers

PLZ HELP, AND PLZ EXPLAIN

shutvik [7]

Answer:

C

Step-by-step explanation:

To make it easy let's start by organizing our information :

AC=12 AND BD=8
ABCD is a rhombus
K and L are the midpoints of sides AD and CD
we notice that the rhombus ABCD is divided into four right triangles

What do you think of when you hear a right triangle ?

The pythagorian theorem !

AC and BD are khown so let's focus on them .

If we concentrated we can notice that AB and BD are cossing each other in the midpoints . why ?

Simply because they are the diagonals of a rhombus .

ow let's apply the pythagorian theorem :

(AC/2)² + (BD/2)² = BC²
6²+4²=52
BC²= 52⇒ $\sqrt{52}$ =BC

Now we khow that : AB=BC=CD=AD= $\sqrt{52}$

This isn't enough . Let's try to figure out a way to calculate the length of KL wich is the base of the triangle

KL is parallel to AC
k is the midpoint of AD and L of DC

I smell something . yes! Thales theorem

KL/AC=DL/DC=DK/AD WE4LL TAKE OLY ONE

KL/12= $\sqrt{52}$ /2* $\sqrt{52}$
KL/12=1/2⇒ KL=6

Now we have the length of the base kl

Now the big boss the height :

notice that you khow the length of KL
BD crosses kl from its midpoint and DL = $\sqrt{52}$ /2

What I want to do is to apply the pythgorian thaorem to khow the lenght of that small part that is not a part of the height of the triangle . I will call it D

DL²=(KL/2)²+D²
52/4= 9+ D²
D² = 52/4-9 +4 SO D=2

now the height of the trigle is H= BD-D= 8-2=6

NOw the area of the triangle is :

A=(KL*H)/2 ⇒ A= (6*6)/2=18

THE ANSWER IS 18 SQ.UN

5 0

3 years ago

Is -9/4 rational. Please I need this now the thing said 20 characters long so I’m just gonna type. The best answer gets brainlie

mihalych1998 [28]

Answer:

no -9/4 is irrational

Step-by-step explanation:

vacsusvvsjshsvsh plz give brainliest

3 0

3 years ago