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Kobotan [32]
3 years ago
6

What is a multiple of 6,9,18 between 61 and 107

Mathematics
1 answer:
Gala2k [10]3 years ago
4 0
There are two possible answers: 72 and 90.

We can find the answer by looking at the largest number among the three of them: 18. The only multiples of 18 between 61 and 107 are 72 and 90. We can check that both numbers are also multiples of 6 and 9:

90 : 6 = 15
90 : 9 = 10

72 : 6 = 12
72 : 9 = 8
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Distrubutive property <br> 1. 54+27=9 (_+_)
Anarel [89]

Answer:

9(6+3) = 54+27

Step-by-step explanation:

dunno how to explain but hope the answer helps

3 0
2 years ago
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What is the x intercept od this piecewise function? A) (0,2) B) (2,0) C) (0,4) D) (4,0)​
Arte-miy333 [17]

Answer: Choice B) (2,0)

======================================

Set the first piece equal to zero and solve for x

x^2 - 4 = 0

x^2 = 4

x = sqrt(4) or x = -sqrt(4)

x = 2 or x = -2

Keep in mind that the first piece y = x^2-4 is only graphed when x = 2 or larger, so we ignore x = -2. This is one x intercept, but there may be more. Let's check the other graph to see what we get.

----------

Set the second piece equal to zero and solve for x

x-2 = 0

x-2+2 = 0+2

x = 2

We get the same result as above in the prior section. Because of this, the two pieces connect at this junction point.

The x intercept x = 2 leads to the location (2,0). The x intercept always occurs when y = 0.

The graph below shows this.

side note: the red piece y = x^2 - 4 looks linear, but it's actually not a straight line. It's just a really stretched out curve.

4 0
3 years ago
What is 985 + 700 + 75 + 10 + 200 - 110?
brilliants [131]

Answer: 1460

Step-by-step explanation: I put it in a calculator.

5 0
2 years ago
The sum of the polynomials 6x3 + 8x2 – 2x + 4 and 10x3 + x2 + 11x + 9 is
faltersainse [42]

Answer:

16x^2 + 9x^2 + 9x + 13.

Step-by-step explanation:

6x^3 + 8x^2 – 2x + 4 +10x^3 + x^2 + 11x + 9

Bringing like terms together:

= 6x^3 + 10x^3 + 8x^2 + x^2 - 2x + 11x + 4 + 9

= 16x^2 + 9x^2 + 9x + 13.  (answer).

4 0
3 years ago
Which expression is equivalent to |a|≤3 ?
weqwewe [10]
<h3>Answer: Choice 2.   a \ge -3 \ \text{ and } \ a \le 3</h3>

=====================================================

Explanation:

The rule is that if |a| \le k for some positive number k, then it can be written as the compound inequality -k \le a \le k. This is a shorthand way of writing the two inequalities a \ge -k \ \text{ and } \ a \le k combined together.

note: -k \le a is the same as a \ge -k

From here, we replace k with 3

Therefore, |a| \le 3 turns into -3 \le a \le 3 which breaks down further to a \ge -3 \ \text{ and } \ a \le 3

The keyword "and" is important as we can't pick one or the other. We must pick both a \ge -3 and also a \le 3

8 0
2 years ago
Read 2 more answers
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