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Misha Larkins [42]
2 years ago
9

The difference between 2 miles and 3 kilometers is how many meters?​

Mathematics
1 answer:
Alchen [17]2 years ago
7 0

Answer:

2 miles = 3218.6m

3km = 3000m

ATQ, 2 miles - 3km = 3218.6m - 3000m =

<em><u>218.6m</u></em>

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What is the area of this figure?
KatRina [158]
Hi!

I have attached 2 images that should help you understand :)

First, look at the edits I made to the image you posted. I separated the shape into smaller shapes so that we can find the area of each individual one. 

Let's start with the rectangle.

To find the area of a rectangle, multiply the width times the height. 
10 · 4 = 40

Rectangle = 40cm

Next up, the red triangles.

I have included another image showing the triangles combined into rectangles. So we can find the area of the triangles just like we would rectangles!

(let me know if you don't understand how I found the width + height of the triangles)

5 · 10 = 50

Red triangles = 50cm

And finally, the green triangles.

8 · 7 = 56

Green triangles = 56cm

Add it all together and you get...

40 + 50 + 56 = 146

The answer to the question is 146cm.

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2 years ago
There are 24 players on a team. Two of every six players were on the team last year. How
JulijaS [17]
8 players were on the team last year
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2 years ago
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Which of these relations on the set {0, 1, 2, 3} are equivalence relations? If not, please give reasons why. (In other words, if
Delicious77 [7]

Answer:

(1)Equivalence Relation

(2)Not Transitive, (0,3) is missing

(3)Equivalence Relation

(4)Not symmetric and Not Transitive, (2,1) is not in the set

Step-by-step explanation:

A set is said to be an equivalence relation if it satisfies the following conditions:

  • Reflexivity: If \forall x \in A, x \rightarrow x
  • Symmetry: \forall x,y \in A, $if x \rightarrow y,$ then y \rightarrow x
  • Transitivity: \forall x,y,z \in A, $if x \rightarrow y,$ and y \rightarrow z, $ then x \rightarrow z

(1) {(0,0), (1,1), (2,2), (3,3)}

(3) {(0,0), (1,1), (1,2), (2,1), (2,2), (3,3)}

The relations in 1 and 3 are Reflexive, Symmetric and Transitive. Therefore (1) and (3) are equivalence relation.

(2) {(0,0), (0,2), (2,0), (2,2), (2,3), (3,2), (3,3)}

In (2), (0,2) and (2,3) are in the set but (0,3) is not in the set.

Therefore, It is not transitive.

As a result, the set (2) is not an equivalence relation.

(4) {(0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,2), (3,3)}

(1,2) is in the set but (2,1) is not in the set, therefore it is not symmetric

Also, (2,0) and (0,1) is in the set, but (2,1) is not, rendering the condition for transitivity invalid.

5 0
3 years ago
Sandy had 10 dollars in her collection. she gave 3 to her younger sister. her parents were so impressed that they quadrupled (le
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Mashutka [201]

\implies {\blue {\boxed {\boxed {\purple {\sf {A. \:11}}}}}}

\sf \bf {\boxed {\mathbb {STEP-BY-STEP\:\:EXPLANATION:}}}

2 \: ( \: x + 1 \: ) - 3

Plugging in the value "x\:=\:6" in the above expression, we have

= 2 \: ( \: 6 + 1 \: ) - 3

= 2 \: ( \: 7 \: ) - 3

= 14 - 3

= 11

<h3><u>Note</u>:-</h3>

\sf\red{PEMDAS\: rule.}

P = Parentheses

E = Exponents

M = Multiplication

D = Division

A = Addition

S = Subtraction

\large\mathfrak{{\pmb{\underline{\orange{Mystique35 }}{\orange{❦}}}}}

5 0
3 years ago
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