All categories

2 years ago

If the markings were missing on this quadrilateral, how would you go about verifying what shape it is?

Mathematics

1 answer:

SVETLANKA909090 [29]2 years ago

7 0

Answer:

Option A: Because this is a rectangle, I would show that each vertex is made up with lines that have opposite reciprocal slopes.

Step-by-step explanation:

A rectangle is a quadrilateral with four right angles.

Each right angle is made up of two perpendicular lines forming a vertex

And then proving this that all perpendicular lines have negative reciprocal slope.

It isn't always necessary to employ the idea of using a parallelogram first although it might be useful.

You might be interested in

Sofia tem 52 triângulos retângulos isósceles iguais. Ela quer fazer quadrados usando alguns desses triângulos. Ela pode fazer qu

oee [108]

Answer:

Portanto, o Sofia pode fazer 6 tamanhos diferentes

Step-by-step explanation:

A fim de sermos capazes de determinar efetivamente quantos quadrados de diferentes tamanhos podem ser formados a partir do triângulo isósceles 52, resolvemos isso usando a fórmula

2 (n + 1) Onde n = 0 e todos os inteiros positivos

em outro para obter um tamanho diferente para os quadrados, temos que adicionar mais triângulos para aumentar o tamanho

a) quando n = 0

2 (0 + 1) = 2 × 1 = 2 triângulos isósceles iguais

A união de 2 triângulos isósceles forma um quadrado. Este é o primeiro quadrado

= 1 quadrado

b) quando n = 1

2 (1 + 1) = 2 × 2 = 4 triângulos isósceles iguais = 1 quadrado

c) quando n = 2

2 (2 + 1) = 2 × 3 = 6 triângulos isósceles iguais = 1 quadrado

d) quando n = 3

2 (3 + 1) = 2 × 4 = 8 triângulos isósceles iguais = 1 quadrado

e) quando n = 4

2 (4 + 1) = 2 × 5 = 10 triângulos isósceles iguais = 1 quadrado

f) quando n = 5

2 (5 + 1) = 2 × 6 = 12 triângulos isósceles iguais = 1 quadrado

Adicionamos o número total de triângulos usados

2 + 4 + 6 + 8 + 10 + 12 = 42 triângulos

42 triângulos de 52 triângulos formariam 6 quadrados com tamanhos diferentes

Vamos fazer mais um, onde n = 6

g) quando n = 6

2 (6 + 1) = 2 × 7 = 14 triângulos isósceles iguais = 1 quadrado

2 + 4 + 6 + 8 + 10 + 12 + 14 = 56 triângulos

Isso já excedeu o número dado de triângulos em questão.

Portanto, o Sofia pode fazer 6 tamanhos diferentes de quadrados usando 42 triângulos isósceles iguais

8 0

2 years ago

Can someone helpp with thissssssssssssssss

Law Incorporation [45]

Answer:

Step-by-step explanation:

let’s say each side length is called a, b, and c

to find out if a side length is a solution you have to know that

a < b + c

b < a + c

c < a + b

if all of these are true then the given lengths can be the sides of a triangle

ex:

a = 3, b = 4, c = 5

3 < 4+5

3 < 9

this is true

4 < 3+5

4 < 8

this is true

5 < 3+4

5 < 7

this is true

so these side lengths can make a triangle

however, if one of these inequalities weren’t true (ex: 9 < 4) you wouldn’t be able to make a triangle with the given side lengths

7 0

2 years ago

By ordering the data find the Lower Quartile<br> 5, 7, 4, 3, 2, 6, 8, 9, 10, 15 *

Klio2033 [76]

Answer:

.Lower quartile = 3.5

Step-by-step explanation:

Rearranging the data from lowest to highest.

2, 3, 4, 5, 6, 7, 8, 9 , 10, 15

First finding the median to make it easier to find the lower quartile.

median : 6 and 7 but right now they're irrelevant so we don't honestly need to work that out.

The lower quartile would be 3 and 4

3+4= 7 divide by 2 = 3.5 is your lower quartile

8 0

2 years ago

I need help plzz and thank u with math

olganol [36]

The answer is 4.81x 10^13. By the way your welcome.

7 0

2 years ago

You have just isolated a new radioactive element. If you can determine its half-life, you will win the Nobel Prize in physics. Y

Troyanec [42]

Staals half
1 gram left
0.3 grams left
half lives are normally written in years
it goes likt this

$A=P( \frac{1}{2})^ \frac{t}{h}$
A=final amount
P=initila amount
t=time elapsed
h=time half life is (same units as time)

let's use days as the units of time

we are given
initial amount=1gram
final amount=0.3gram
time=4 days

so
$0.3=1( \frac{1}{2})^ \frac{4}{h}$
solve for h
$0.3=( \frac{1}{2})^ \frac{4}{h}$
take the ln of both sides
$ln(0.3)=ln( \frac{1}{2})^ \frac{4}{h}$
$ln(0.3)=\frac{4}{h}ln( \frac{1}{2})$
times both sides by h
$hln(0.3)=4ln( \frac{1}{2})$
divide both sides by ln(0.3)
$h= \frac{4ln( \frac{1}{2}) }{ln(0.3)}$
use calculator
h=2.30287
3 sig figs
h=2.303 days

6 0

3 years ago