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gregori [183]
3 years ago
7

Use the number line to determine the unknown addend in the given number sentence.

Mathematics
2 answers:
Gnesinka [82]3 years ago
5 0

Answer:

C

Step-by-step explanation:

-4+(-5)=-9

-4=-9+5

-4=-4......correct

kotykmax [81]3 years ago
4 0
The answer would be c.-5
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Which equations show direct variation between a and b?
ipn [44]
Formula of direct variation is: y = kx 
k is the constant of variation
y is the total value that is directly varies with the changes in x.

in this case. x is represented by a. y is represented by b.

A.) b = 0.7a  ⇒DIRECT VARIATION with 0.7 as the constant of variation

B.) b = 1/8 a⁻³  ⇒ Not a direct variation. Because a has a negative exponent

C.) 5(1/a) = b ⇒ Not a direct variation

D.) 1a = b ⇒ DIRECT VARIATION with 1 as the constant of variation

6 0
3 years ago
Pick the decimal equivalent to 3 9/100
Sidana [21]
If you have pick it, then you should be offered choices and you should include them in the question.

The answer is 3.09
4 0
3 years ago
Read 2 more answers
Simplify (2x^4 + 7x^3 - 3x^2 + 7) - (4x^4 - 3x^3 - 5x - 20)
nikdorinn [45]

Answer:

  -2x^4 +10x^3 -3x^2 +5x +27

Step-by-step explanation:

a) Using the distributive property, we can eliminate parentheses:

  2x^4 + 7x^3 - 3x^2 + 7 - 4x^4 + 3x^3 + 5x + 20

And using the distributive property again, we can factor pairs of like terms:

  = x^4(2 -4) +x^3(7 +3) -3x^2 +5x +(7 +20)

b)

  = -2x^4 +10x^3 -3x^2 +5x +27

4 0
3 years ago
Read 2 more answers
The table shows the scores for three teams over five games.​
krek1111 [17]

The team with mode with the highest score is team C. Because mode is the number you see more than once.And the mode for Team A is 56, Team B is 60, and Team C is 82.

7 0
3 years ago
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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