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nydimaria [60]
3 years ago
15

[P(x) + Q(x)] · R(x)

Mathematics
1 answer:
Mademuasel [1]3 years ago
5 0

r(X(5+4) YOUR now welcome

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Please help! If you can answer both that would be great, thank you!!!
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I cant see the numerators on the numbers on the top one but the last one is 53/60.

7 0
3 years ago
1.Give one number that makes the inequality x + 7 > 20 true
nata0808 [166]
1. Any number above 13 works. Why? Because 20-7=13, and to be greater than 20, you must add a number larger than 13.

Examples: 14+7 > 20, 30+7 > 20, 100+7 > 20

2. Any number below 25/3 (which is also 8.3 with a repeating 3) works. Why? Because 25/3=8.3 with a repeating 3, and to remain less than 25, you must multiply by a number less than 8.3 with a repeating 3.

Examples: 3(8) < 25, 3(5) < 25, 3(0) < 25

3. 4 buses. 1 bus will hold 60 students, 2 will hold 120, 3 will hold 180, and 4 will hold 240. The question is trying to trick you into putting now 3.3333333333... buses because that's what 200/60 is, but there is no such thing as a third of a bus. So you need at least 4 buses. (There will be an extra 40 spaces for passengers on the 4th bus, but that is okay.)

To find this answer I did 200/60 and got 3.3 with a repeating 3. You must round to the higher whole number. Rounding down to 3 buses leaves you with 20 students without a bus.

4. 19 boxes. 18 boxes will only hold 288 candies. The question is trying to trick you into putting down 18.75 boxes because that's what 300/16 is, but there is no such thing as 75% of a box. So you need at least 19 boxes. (There will be an extra 4 spaces for candies in the 19th box, but that is okay.)

To find this answer I did 300/16 and got 18.75. You must round to the higher whole <span>number. Rounding down to 18 boxes leaves you with 12 candies without a box.</span>
6 0
3 years ago
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