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Vitek1552 [10]
3 years ago
14

Angle a and angle b are both supplementary angles. What must be true about the angles?

Mathematics
1 answer:
melamori03 [73]3 years ago
3 0

Not sure if there're any answer choices of some sort,

but if the angle a and angle b are supplementary, that means when you add their measures together they must equal 180.


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Step-by-step explanation:

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For which of the following compound inequalities is there no solution? ​
Bess [88]

Answer:

(a)6m \le - 36 and m + 24 > 20

Step-by-step explanation:

Required

Which of a to d has no solution

(a)6m \le - 36 and m + 24 > 20

We have: m + 24 > 20

Sole for m

m >20 - 24

m >-4

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Divide by 6

m \le -6

So, we have:

m \le -6 and m >-4

m \le -6 implies that: m = -6,-7,-8,-9.....

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<em>Hence, there is no solution</em>

8 0
3 years ago
Find the HCF of 27, 63, 54 using prime factorization method.
Step2247 [10]

Answer:

27 = 3 , 3, 3

63= 3 , 3, 7

54= 3 , 3, 3, 2

HCF OF 27, 63 AND 54 = 3 X 3 = 9

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7 0
3 years ago
There are 8 members on a board of directors. if they must form a subcommittee of 6 members, how many different subcommittees are
Shtirlitz [24]
1. Given a group of n people. There are C(n, r) ways of forming groups of r out of n.

2. Where C(n, r)=\frac{n!}{r!(n-r)!}

3. For example, given {Andy, John, Julia}. We want to pick 2 people to give a gift: we can pick {(Andy, John), (Andy, Julia), (John, Julia)}, so there are 3 ways. So we can list and count.

4. Or we could do this with the formula C(3, 2)=\frac{3!}{2!(1)!}= \frac{3*2*1}{2*1}=3

5. C(8, 6)=\frac{8!}{2!6!}= \frac{8*7*6!}{2*6!}= \frac{8*7}{2}= 4*7=28

So there are C(8,6)=28 ways of chosing 6 out of 8 people to form the subcommittees. <span />
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There are two numbers, and one is 7 more than twice the other. The sum of the numbers is 43. What are the numbers? If x is less
chubhunter [2.5K]

Answer:

43-7=x

Step-by-step explanation:

5 0
2 years ago
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