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labwork [276]
3 years ago
5

Help: The ratio of areas of two similar figures is 8:18. The perimeter of the larger figure is 24, find the perimeter of the sma

ller figure.\
THXXX :
Mathematics
1 answer:
Georgia [21]3 years ago
3 0

Lets use proportion

let the perimeter of the smaller figure be 'x'

8:18::x:24

Product of means=product of extreames

24 x 8= 18 x 'x'

192=18x

x=10.6666667 units

Ans: the perimeter of the smaller figure is 10.6666667 units

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Simplify √9 x 3√27 please answer
irakobra [83]

Answer:

I got 27\sqrt{3} which is 46.76537.... in decimal form.

Step-by-step explanation:

4 0
3 years ago
N =⟨−2,  −1⟩ and D=<img src="https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-3%260%5C%5C4%261%5C%5C%5Cend%7Barray%
zepelin [54]

Answer:

6

-9

Step-by-step explanation:

-3(-2) + 0(-1)

4(-2) + 1(-1)

6 + 0

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*don't forget to add brackets on each side of every calculation okay :)

5 0
3 years ago
In ABC, side BC is extended to point E. When connected to vertex A, segment EA is parallel to segment BD. In this construction,
Anuta_ua [19.1K]

Two triangles are said to be <u>congruent</u> if they have <em>similar</em> properties. Thus the required <u>options</u> to complete the <em>paragraph proof</em> are:

a. angle 1 is <u>congruent</u> to angle 2.

b. <em>alternate</em> angles are <u>congruent</u> if two parallel lines are cut by a <em>transversal</em>.

c. \frac{AD}{CD} = \frac{AB}{CB}

The <em>similarity property</em> of two or more shapes implies that the <u>shapes</u> are congruent. Thus they have the <em>same</em> properties.

From the given <u>diagram</u> in the question, it can be deduced that

ΔABC ≅ ΔABE (<em>substitution</em> property of equality)

Given that EA is <u>parallel</u> to BD, then:

i. <2 ≅ <3 (<em>corresponding</em> angle property)

ii. <1 ≅ < 4 (<em>alternate</em> angle property)

Thus, the required options to complete the <em>paragraph proof</em> are:

  • Angle 1 is <em>congruent</em> to angle 2.
  • Alternate angles are <u>congruent</u> if two parallel lines are cut by a <em>transversal</em>.
  • \frac{AD}{CD} = \frac{AB}{CB}

For more clarifications on the properties of congruent triangles, visit: brainly.com/question/1619927

#SPJ1

6 0
2 years ago
Extension question (provide a full explanation of your method(s):
Volgvan

Answer:

Ann has little chance to win if she is presented with 4 counters.

Ann can always win from a pile of 6 counters.

(both are explained below)

Step-by-step explanation:

If Ann  is presented with 4 counters, and

1. if she takes out 3, she will lose since the opponent will  pull out 1 and the last one.

2. if she takes 2 her opponent will take out 1 and she can't pull out the last 1 since her opponents last move was to pull out 1  counter so she will lose.

3. If she takes out 1 and her opponent takes out 3 in the next move she loses.

but if instead of 3 her opponent takes out 2 and in the last move Ann takes out the last 1  then she will win.

So, If Ann is presented with 4 counters she has little chance to win provided in the move just before, her opponent didn't move 1 counter.

Now,

if there is 6 counters to Ann, and

1., if Ben's  previous move was 1 then Ann can win if she takes out 3 or 2.

If she takes out 3 Ben can take out 1 or 2 and in the last move she will take out 2 or 1 (respectively) and winning the game.

If she takes out 2 Ben can  take out 1 or 3 and in the last move Ann wins by pulling out 3 or 1 respectively.

2. if Ben's  previous move was 2 then Ann can win if she takes out 1 or 3.

If she takes out 1 Ben can take out 2 or 3 and in the last move she will take out 3 or 2(respectively) and winning the game.

If she takes out 3 Ben can  take out 1 or 2 and in the last move Ann wins by pulling out 2 or 1 respectively.

2. if Ben's  previous move was 3 then Ann can win if she takes out 1 or 2.

If she takes out 1 Ben can take out 2 or 3 and in the last move she will take out 3 or 2(respectively) and winning the game.

If she takes out 2 Ben can  take out 1 or 3 and in the last move Ann wins by pulling out 3 or 1 respectively.

 

 

7 0
4 years ago
[ÑL;<br>´lkCuáles son las cosas que pueden medir en fracciones y cuáles en decimales
Vlada [557]

Answer:

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Step-by-step explanation:

Las cosas que se pueden contar (i.e. Contar y clasificar frutas) así como cosas que se pueden dividir se pueden medir en fracciones (i.e. Cortar un pastel), mientras que las cosas que no se pueden contar, de naturaleza no discreta, así como cosas que no se pueden dividir (i.e. Líquidos, material granulado) se pueden medir en decimales.

8 0
4 years ago
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