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Kryger [21]
3 years ago
13

How can the statement be rewritten as a conditional statement in if-then form? A rectangle with 4 congruent sides is a square.

Mathematics
2 answers:
Nady [450]3 years ago
7 0

Answer:

it is a

Step-by-step explanation:

larisa86 [58]3 years ago
3 0

Answer:

the answer is a

Step-by-step explanation:

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1/36

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There's only one situation where you can roll a sum of 2 (1,1)

so the final answer is just 1/36

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Read 2 more answers
Afla un numar,stiind ca: a)jumatatea treimii lui este 25; b)sfertul jumatatii lui este 48; c)treimea sfertului sau,marita cu 17
aalyn [17]

Answer:

(a) 150

(b) 384

(c)  84

(d) 261

(e ) 480

Step-by-step explanation:

(a)

Let 'a' be a number such that  the half of his third is 25; i.e.

\dfrac{1}{2}\times(\dfrac{1}{3}a)=25\\\\\dfrca{1}{2\times3}a=25\\\\\dfrac{1}{6}\times a=25\\\\a=25\times6\\\\a=150    

Hence, the number is 150.

(b)

Let 'b' be  the number such that the quarter of his half is 48 i.e.

\dfrac{1}{4}(\dfrac{1}{2}b)=48\\\\\dfrac{1}{8}b=48\\\\b=48\times8\\\\b=384

Hence the number is 384.

(c)

Let 'c' be the number such that the third quarter or, increased by 17 is 80.

\dfrac{3}{4}c+17=80\\\\\dfrac{3}{4}c=80-17\\\\\dfrac{3}{4}c=63\\\\c=\dfrac{63\times4}{3}\\\\c=\dfrac{252}{3}\\\\c=84

Hence, the number is 84.

(d)

Let 'd'  be the number such that its triplet, reduced by 28 is 755

3d-28=755\\\\3d=755+28=783\\\\d=\dfrac{783}{3}\\\\d=261

Hence the number is 261.

(e  

Let 'e' be the number such that double his third, increased by 80 is 400.

\\2\dfrac{1}{3}e+80=400\\\\\dfrac{2}{3}e=400-80=320\\\\e=\dfrac{320\times 3}{2}\\\\e=\dfrac{960}{2}\\\\e=480

Hence, the value of the number is 480.


4 0
4 years ago
I need help plzzzzz help im time so i need it quick plzzzz
Gnom [1K]
<h3> - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - </h3>

➷ You have to use pythagoras' theorem to find the diameter.

c^{2} = a^{2} +b^{2}

c^{2} =6^{2} +6^{2}

c^{2} =72

c=\sqrt{72}

➶Hope This Helps You!

➶Good Luck :)

➶Have A Great Day ^-^

↬ Hannah ♡

8 0
3 years ago
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