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Juliette [100K]
3 years ago
6

Which expression has the same value as the one below?10 +(-3)​

Mathematics
2 answers:
Sergio039 [100]3 years ago
6 0
10-3 is the same as 10 +(-3)
Anna35 [415]3 years ago
4 0
10+ (-3)=10-3=7 is the answer
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20.35 divided by 3 equals 6 (:
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PLZ HURRY IT'S URGENT!!
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Answer: C. 375

Step-by-step explanation:

All your doing is multiplying

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3 years ago
Complete the proof of the exterior angle theorem. Given: ∠4 is an exterior angle of △ABC. Prove: m∠1+m∠2=m∠4 Answer: Statement R
Nataly_w [17]

Answer:


Step-by-step explanation:

Statement                                                                              Reason

1. ∠4 is an exterior angle of ΔABC                                         Given

2. ∠3 and ∠4 form a linear pair                                    Linear pair property

3. ∠3 is supplementary to ∠4                                     Straight line property

4. m∠3+m∠4=180°                                                ∠3 is supplementary to ∠4      

5. m∠1+m∠2+m∠3=180°                                           Triangle sum theorem

6.m∠1+m∠2+m∠3= m∠3+m∠4                                   Statement 4 and 5

7. m∠1+m∠2=m∠4                                                     Subtraction property

5 0
3 years ago
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Fran is limited to watching television less than 12.6 hours per week. She has already watched 4.2 hours, and each show is 0.7 of
malfutka [58]

Answer: the number of hpurs that Fran can watch this week is lesser than 12

Step-by-step explanation:

Let x represent the number of hours of television that Fran can watch in a week.

Fran is limited to watching television less than 12.6 hours per week. She has already watched 4.2 hours, and each show is 0.7 of an hour long. The inequality representing the situation is expressed as

0.7x + 4.2 < 12.6

0.7x + 4.2 < 12.6 - 4.2

0.7x < 8.4

x < 8.4/0.7

x < 12

4 0
3 years ago
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How to get the answers?​
beks73 [17]

By inclusion/exclusion,

n(P' \cup Q) = n(P') + n(Q) - n(P' \cap Q)

We have

n(\xi) = n(P) + n(P') \implies n(P') = 28 - n(P)

so that

n(P'  \cup Q) = (28 - n(P)) + n(Q) - 2n(P) = 28 - 3n(P) + n(Q)

Now,

n(\xi) = 28 \implies n(P \cup Q) = 28 - 7 = 21

and by inclusion/exclusion,

n(P \cup Q) = n(P) + n(Q) - n(P \cap Q)

Decompose Q into the union of two disjoint sets:

Q = (P \cap Q) \cup (P' \cap Q)

Since they're disjoint,

n(Q) = n(P\cap Q) + n(P'\cap Q) \implies n(Q) = n(P\cap Q) + 2n(P)

\implies n(P \cup Q) = n(P) + (n(P\cap Q) + 2n(P)) - n(P \cap Q)

\implies 21 = 3n(P)

\implies n(P) = 7

From the Venn diagram, we see there are 3 elements unique to P - by the way, this is the set P \cap Q' - so n(P\cap Q) = 7-3 = 4, and it follows that

n(Q) = n(P\cap Q) + 2n(P) = 4 + 2\times7 = 18

Finally, we get for (a)

n(P' \cup Q) = 28 - 3n(P) + n(Q) = 28 - 3\times7 + 18 = \boxed{25}

For (b), we have by inclusion/exclusion that

n(P \cup Q') = n(P) + n(Q) - n(P \cap Q') = 7 + 18 - 3 = \boxed{22}

5 0
2 years ago
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